Photoelectric emission is observed from a surface for frequency ${v_1}$ and ${v_2}$ of the incident radiation $\left( {{{\rm{v}}_1} > {{\rm{v}}_2}} \right).$ If the maximum kinetic energies of the photoelectrons in the two cases are in the ratio 1∶k, then the threshold frequency ${{\rm{v}}_0}$ is given by:

$\frac{{{\rm{k}}{{\rm{v}}_1} - {{\rm{v}}_2}}}{{k - 1}}$

$\frac{{{{\rm{v}}_2} - {{\rm{v}}_1}}}{{k - 1}}$

$\frac{{{\rm{k}}{{\rm{v}}_2} - {{\rm{v}}_1}}}{{k - 1}}$

$\frac{{{{\rm{v}}_2} - {{\rm{v}}_1}}}{k}$

The threshold frequency of a metal is $4 \times {10^{14}}{{\rm{s}}^{ - 1}}.$. The minimum energy of photon to cause photoelectric effect is:

$2.64 \times {10^{ - 19}}{\rm{\;J}}$

$3.06 \times {10^{ - 12}}{\rm{\;J}}$

$1.4 \times {10^{ - 18}}{\rm{\;J}}$

$3.4 \times {10^{ - 19}}{\rm{\;J}}$

Given : The mass of electron is $9.11 \times {10^{ - 31}}{\rm{kg}}$ and Planck constant is $6.626 \times {10^{ - 34}}{\rm{\;Js}},$ the uncertainty involved in the measurement of velocity within a distance of $0.1\;$ is:

$5.79 \times {10^6}{\rm{\;m\;}}{{\rm{s}}^{ - 1}}$

$5.79 \times {10^8}{\rm{\;m\;}}{{\rm{s}}^{ - 1}}$

$5.79 \times {10^5}{\rm{\;m\;}}{{\rm{s}}^{ - 1}}$

$5.79 \times {10^7}{\rm{\;m\;}}{{\rm{s}}^{ - 1}}$

The uncertainty in the momentum of an electron is $1.0 \times {10^{ - 5}}{\rm{kg\;m}}{{\rm{s}}^{ - 1}}$ , The uncertainty in its position will be

$5.27 \times {10^{ - 30}}{\rm{m}}$

$1.50 \times {10^{ - 28}}{\rm{m}}$

$1.05 \times {10^{ - 26}}{\rm{m}}$

$5.25 \times {10^{ - 28}}{\rm{m}}$

The ionisation enthalpy of hydrogen atom is $1.312 \times {10^6}{\rm{Jmo}}{{\rm{l}}^{ - 1}}$. The energy required to excite the electron in the atom from ${n_1} = 1\;{\rm{to }}\;{n_2} = 2$ is

$9.84 \times {10^5}{\rm{\;J\;mo}}{{\rm{l}}^{ - 1}}$

$8.51 \times {10^{5{\rm{\;}}}}{\rm{J\;mo}}{{\rm{l}}^{ - 1}}$

$6.56 \times {10^5}{\rm{\;J\;mo}}{{\rm{l}}^{ - 1}}$

$7.56 \times {10^5}{\rm{\;J\;mo}}{{\rm{l}}^{ - 1}}$

STRUCTURE OF ATOM MCQs for NEET UG CHEMISTRY — Hard, Page 2

31 Hard

31 of 97Question permalink

Application +4 | -1

Photoelectric emission is observed from a surface for frequency

{v_1}

and

{v_2}

of the incident radiation

\left( {{\mathrm{v}_1} > {\mathrm{v}_2}} \right).

If the maximum kinetic energies of the photoelectrons in the two cases are in the ratio 1∶k, then the threshold frequency 

{\mathrm{v}_0}

 is given by:

View answer and explanation

Correct option: B

\frac{{\mathrm{k}{\mathrm{v}_1} - {\mathrm{v}_2}}}{{k - 1}}

Option B (

\frac{{\mathrm{k}{\mathrm{v}_1} - {\mathrm{v}_2}}}{{k - 1}}

) is correct because it represents the threshold frequency derived from the photoelectric effect equations and the given ratio of kinetic energies.

Solution approach

Correct Answer + Core Reason: $\frac{{\mathrm{k}{\mathrm{v}_1} - {\mathrm{v}_2}}}{{k - 1}}$ . This is the threshold frequency derived from the photoelectric effect equations and the given ratio of kinetic energies.

Step-by-Step Solution:

Step 1: Write the photoelectric equation for both frequencies:
$K{E_1} = h{v_1} - h{v_0}$
$K{E_2} = h{v_2} - h{v_0}$

Step 2: Given that $\frac{{K{E_1}}}{{K{E_2}}} = \frac{1}{k}$ , so $K{E_2} = kK{E_1}$

Step 3: Substitute $K{E_2}$ in the second equation:\nkK{E_1} = h{v_2} - h{v_0} $\n\n**Step 4:** Substitute$ K{E_1} = h{v_1} - h{v_0} $:\nk(h{v_1} - h{v_0}) = h{v_2} - h{v_0}$

Step 5: Expand and rearrange to solve for ${v_0}$ :\nkh{v_1} - kh{v_0} = h{v_2} - h{v_0} $\nkh{v_1} - h{v_2} = kh{v_0} - h{v_0}$ \nh(k{v_1} - {v_2}) = h{v_0}(k - 1) $\n$ {v_0} = \frac{{k{v_1} - {v_2}}}{{k - 1}} $\n\n**Why other options are wrong:**\n- **Option A:**$ \frac{{{{v}_2} - {{v}_1}}}{{k - 1}} $: This is incorrect because it subtracts the frequencies in the wrong order and doesn't account for the factor 'k' correctly.\n- **Option C:**$ \frac{{{k}{{v}_2} - {{v}_1}}}{{k - 1}} $: This is incorrect because it swaps$ {v_1} $and$ {v_2} $in the numerator.\n- **Option D:**$ \frac{{{{v}_2} - {{v}_1}}}{k} $: This is incorrect because it doesn't account for the factor 'k' correctly and the subtraction is in the wrong order.\n\n**Key Concepts:**\n- Photoelectric effect\n- Threshold frequency\n- Kinetic energy of photoelectrons\n\n**Memory Aids:**\n- Photoelectric equation:$ KE = hν - φ $, where$ φ = hν_0 $\n\n**Quick Tips:**\n- Always start with the photoelectric equation.\n- Pay attention to the ratio of kinetic energies.$

32 Hard

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Analyzing +4 | -1

According to Bohr’s atomic theory,
I. kinetic energy of electron is

\propto \,\,\frac{{{Z^2}}}{{{n^2}}}

II. the product of velocity

(v)\,

of electron and principal quantum number

(n),\,\,'vn' \propto {Z^2}

.
III. frequency of revolution of electron in an orbit is

\propto \frac{{{z^3}}}{{{n^3}}}

IV. coulombic force of attraction on the electron is

\propto \frac{{{z^3}}}{{{n^4}}}

Choose the most appropriate answer from the options given below.

View answer and explanation

Correct option: D

I and IV

Option D (I and IV) is correct. The kinetic energy is proportional to

Z^2/n^2

and the coulombic force is proportional to

Z^3/n^4

.

Solution approach

Correct Answer + Core Reason: I and IV. These statements correctly relate the kinetic energy and coulombic force to Z and n.

Step-by-Step Solution:

Step 1: Recall Bohr's atomic theory postulates and related formulas.

Step 2: Analyze each statement:
I. Kinetic energy (KE) of electron: $KE = \frac{1}{2}mv^2$ . Since $v \propto \frac{Z}{n}$ , $KE \propto \frac{Z^2}{n^2}$ . This statement is correct.
II. Product of velocity (v) and principal quantum number (n): $v \propto \frac{Z}{n}$ , so $vn \propto Z$ . The given statement says $vn \propto Z^2$ , which is incorrect.
III. Frequency of revolution (f): $f = \frac{v}{2\pi r}$ . Since $v \propto \frac{Z}{n}$ and $r \propto \frac{n^2}{Z}$ , $f \propto \frac{Z}{n} \times \frac{Z}{n^2} \propto \frac{Z^2}{n^3}$ . The given statement says $f \propto \frac{Z^3}{n^3}$ , which is incorrect.
IV. Coulombic force (F): $F = \frac{kZe^2}{r^2}$ . Since $r \propto \frac{n^2}{Z}$ , $F \propto \frac{Z}{(\frac{n^2}{Z})^2} \propto \frac{Z^3}{n^4}$ . This statement is correct.

Step 3: Identify the correct statements. Statements I and IV are correct.

Why other options are wrong:
- Option A: Only III is incorrect because statement III is false.
- Option B: Only I is incorrect because statement IV is also true.
- Option C: I, III, and IV is incorrect because statement III is false.

Key Concepts:
- Bohr's atomic theory
- Kinetic energy of electron
- Frequency of revolution
- Coulombic force

Memory Aids:
- Remember the relationships: $v \propto \frac{Z}{n}$ , $r \propto \frac{n^2}{Z}$

Quick Tips:
- Use the basic formulas and relationships from Bohr's theory to derive the proportionality for each quantity.

33 Hard

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Application +4 | -1

The Vividh Bharti Station of All India Radio, Delhi, broadcasts on a frequency of

1,368 \mathrm{kHz}

(Kilohertz). Calculate the wavelength

(\lambda)

of the electromagnetic radiation emitted by transmitter. Which part of the electromagnetic spectrum does it belong to?

View answer and explanation

Correct option: D

219.3 \mathrm{~m}

and radiowave

Option D (219.3 m and radiowave) is correct because the wavelength is calculated using c = f*lambda and this wavelength falls within the radiowave region of the electromagnetic spectrum.

Solution approach

Correct Answer + Core Reason: $219.3 \mathrm{~m}$ and radiowave. The wavelength is calculated using $c = f\lambda$ and falls in the radiowave region.

Step-by-Step Solution:

Step 1: Identify the given information: frequency (f) = 1368 kHz = 1368 × 10³ Hz, speed of light (c) = 3.0 × 10⁸ m/s.

Step 2: Recall the formula relating wavelength, frequency, and speed of light: $c = f\lambda$ , where $\lambda$ is the wavelength.

Step 3: Rearrange the formula to solve for wavelength: $\lambda = \frac{c}{f}$

Step 4: Substitute the given values into the formula: $\lambda = \frac{{3.0 \times 10^8 \mathrm{~m} / \mathrm{s}}}{{1368 \times 10^3 \mathrm{~Hz}}}$

Step 5: Calculate the wavelength: $\lambda = \frac{{3.0 \times 10^8}}{{1368 \times 10^3}} \mathrm{~m} = 219.29 \mathrm{~m} \approx 219.3 \mathrm{~m}$

Step 6: Determine the region of the electromagnetic spectrum: A wavelength of 219.3 m falls within the radiowave region.

Why other options are wrong:
- **Option A ( $319.4 \mathrm{~m}$ and X-rays): The wavelength calculation is incorrect, and X-rays have much shorter wavelengths.
- Option B ( $319.4 \mathrm{~m}$ and radiowave): The wavelength calculation is incorrect.
- Option C ( $219.3 \mathrm{~m}$ and microwave): The wavelength is correct, but this wavelength falls within the radiowave region, not the microwave region.

Key Concepts:**
- Relationship between speed of light, frequency, and wavelength: $c = f\lambda$
- Electromagnetic spectrum

Memory Aids:
- Remember the order of the electromagnetic spectrum (Radio, Microwave, Infrared, Visible, Ultraviolet, X-ray, Gamma).

Quick Tips:
- Convert frequency to Hz before calculating wavelength.
- Know the approximate wavelength ranges for different regions of the electromagnetic spectrum.

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Analyzing +4 | -1

When photons of energy

4.25 \mathrm{~eV}

strike the surface of a metal

{A}

, the ejected photoelectrons have maximum kinetic energy,

{T}_{{A}}

(expressed in

\mathrm{eV}

) and de-Broglie wavelength

\lambda_{{A}}

. The maximum kinetic energy of photo electrons liberated from another metal

{B}

by photons of energy

4.20 \mathrm{~eV}

is

T_B=\left(T_A-1.50\right) \mathrm{eV}

. If the de-Broglie wavelength of these photo electrons is

\lambda_{{B}}=2 \lambda_{{A}}

, then

View answer and explanation

Correct option: C

{T}_{{A}}=2.00 \mathrm{~eV}

Option C (

{T}_{{A}}=2.00 \mathrm{~eV}

) is correct because by using the de-Broglie wavelength relationship and the photoelectric effect equation, we can solve for the kinetic energy of the ejected photoelectrons from metal A.

Solution approach

Correct Answer + Core Reason: ${T}_{{A}}=2.00 \mathrm{~eV}$ . By using the de-Broglie wavelength relationship and the photoelectric effect equation, we can solve for the kinetic energy of the ejected photoelectrons from metal A.

Step-by-Step Solution:

Step 1: Define the given information:
$E_A = 4.25 \mathrm{~eV}$ , $E_B = 4.20 \mathrm{~eV}$
$T_B = T_A - 1.50 \mathrm{~eV}$
$\lambda_B = 2 \lambda_A$

Step 2: Use the de-Broglie wavelength relationship: $\lambda = \frac{h}{\sqrt{2mK}}$ , where $K$ is the kinetic energy.\nSince $\lambda_B = 2 \lambda_A$ , we have $\frac{h}{\sqrt{2mT_B}} = 2 \frac{h}{\sqrt{2mT_A}}$ .\nThis simplifies to $\sqrt{T_A} = 2 \sqrt{T_B}$ , or $T_A = 4T_B$ .

Step 3: Substitute $T_B = T_A - 1.50$ into $T_A = 4T_B$ :\nT_A = 4(T_A - 1.50)\nT_A = 4T_A - 6
3T_A = 6\nT_A = 2.00 ~eV $\n\n**Step 4:** Calculate$ T_B $:\nT_B = T_A - 1.50 = 2.00 - 1.50 = 0.50 \mathrm{~eV}$

Step 5: Calculate the work functions:
$\phi_A = E_A - T_A = 4.25 - 2.00 = 2.25 \mathrm{~eV}$
$\phi_B = E_B - T_B = 4.20 - 0.50 = 3.70 \mathrm{~eV}$

Why other options are wrong:
- Option A: The work function of ${A}$ is $2.05 \mathrm{~eV}$ . Incorrect, the work function of A is 2.25 eV.
- Option B: The work function of ${B}$ is $3.20 \mathrm{~eV}$ . Incorrect, the work function of B is 3.70 eV.
- Option D: ${T}_{{B}}=2.75 \mathrm{~eV}$ . Incorrect, the kinetic energy of B is 0.50 eV.

Key Concepts:
- Photoelectric effect
- de-Broglie wavelength
- Work function
- Kinetic energy of photoelectrons

Memory Aids:
- $E = hf = \phi + KE$ (Photoelectric equation)
- $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mKE}}$ (de-Broglie wavelength)

Quick Tips:
- Remember to relate the de-Broglie wavelength to the kinetic energy of the photoelectrons.

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35 of 97Question permalink

Application +4 | -1

The threshold frequency of a metal is

4 \times {10^{14}}{\mathrm{s}^{ - 1}}.

. The minimum energy of photon to cause photoelectric effect is:

View answer and explanation

Correct option: D

2.64 \times {10^{ - 19}}\mathrm{\;J}

Option D (

2.64 \times {10^{ - 19}}\mathrm{\;J}

) is correct because the minimum energy is the product of Planck's constant and the threshold frequency.

Solution approach

Correct Answer + Core Reason: $2.64 \times {10^{ - 19}}\mathrm{\;J}$ . The minimum energy is the product of Planck's constant and the threshold frequency.

Step-by-Step Solution:

Step 1: The minimum energy of a photon to cause the photoelectric effect is given by:
$E = h \nu_0$ \nwhere h is Planck's constant and $\nu_0$ is the threshold frequency.

Step 2: Planck's constant (h) is approximately $6.626 \times 10^{-34} \text{ J s}$ . The threshold frequency ( $\nu_0$ ) is given as $4 \times 10^{14} \text{ s}^{-1}$ .

Step 3: Substitute the values into the formula:
$E = (6.626 \times 10^{-34} \text{ J s}) (4 \times 10^{14} \text{ s}^{-1})$
$E = 2.6504 \times 10^{-19} \text{ J}$

Step 4: Therefore, the minimum energy is approximately $2.64 \times 10^{-19} \text{ J}$ .

Why other options are wrong:
- Option A: Incorrect calculation.
- Option B: Incorrect calculation.
- Option C: Incorrect calculation.

Key Concepts:
- Photoelectric effect
- Threshold frequency
- Planck's constant

Memory Aids:
- $E = h \nu_0$ (Minimum energy for photoelectric effect).

Quick Tips:
- Remember the value of Planck's constant and the formula for the minimum energy required for the photoelectric effect.

36 Hard

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Application +4 | -1

Given : The mass of electron is

9.11 \times {10^{ - 31}}\mathrm{kg}

and Planck constant is

6.626 \times {10^{ - 34}}\mathrm{\;Js},

the uncertainty involved in the measurement of velocity within a distance of

0.1\;

is:

View answer and explanation

Correct option: C

5.79 \times {10^6}\mathrm{\;m\;}{\mathrm{s}^{ - 1}}

Option C (

5.79 \times {10^6}\mathrm{\;m\;}{\mathrm{s}^{ - 1}}

) is correct because it represents the uncertainty in velocity calculated using the Heisenberg uncertainty principle.

Solution approach

Correct Answer + Core Reason: $5.79 \times {10^6}\mathrm{\;m\;}{\mathrm{s}^{ - 1}}$ . The uncertainty in velocity is calculated using Heisenberg's uncertainty principle.

Step-by-Step Solution:

Step 1: State Heisenberg's uncertainty principle: $\Delta x \Delta p \geq \frac{h}{4\pi}$ , where $\Delta x$ is the uncertainty in position, $\Delta p$ is the uncertainty in momentum, and $h$ is Planck's constant.

Step 2: Relate momentum to velocity: $\Delta p = m \Delta v$ , where $m$ is the mass and $\Delta v$ is the uncertainty in velocity.

Step 3: Substitute into the uncertainty principle: $\Delta x (m \Delta v) \geq \frac{h}{4\pi}$

Step 4: Solve for $\Delta v$ : $\Delta v \geq \frac{h}{4\pi m \Delta x}$

Step 5: Substitute the values: $h = 6.626 \times 10^{-34} \text{ Js}$ , $m = 9.11 \times 10^{-31} \text{ kg}$ , $\Delta x = 0.1 \times 10^{-6} \text{ m}$
$\Delta v \geq \frac{6.626 \times 10^{-34}}{4 \pi (9.11 \times 10^{-31}) (0.1 \times 10^{-6})} \approx 5.79 \times 10^6 \text{ m/s}$

Step 6: Final Answer: $\Delta v \approx 5.79 \times 10^6 \text{ m/s}$

Why other options are wrong:
- Option A: This is incorrect as it's significantly higher than the calculated value.
- Option B: This is incorrect as it's significantly lower than the calculated value.
- Option D: This is incorrect as it's higher than the calculated value.

Key Concepts:
- Heisenberg's uncertainty principle
- Uncertainty in position and momentum

Memory Aids:
- Remember the Heisenberg uncertainty principle formula.

Quick Tips:
- Ensure consistent units throughout the calculation.
- Pay attention to the given uncertainty in position.

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Understanding +4 | -1

If n=6, the correct sequence for filling of electrons will be:

View answer and explanation

Correct option: A

ns \to np \to \left( {n - 1} \right)d \to \left( {n - 2} \right)f

The question is flawed. The correct sequence for filling electrons when n=6 is based on (n+l) rule and Aufbau principle, which gives the order: 6s → 4f → 5d → 6p. Therefore, option B is the closest to the correct answer.

Solution approach

Correct Answer + Core Reason: $ns \to (n - 2)f \to (n - 1)d \to np$ . This order follows the (n+l) rule and Aufbau principle, although the question has an error in the options.

Step-by-Step Solution:

Step 1: State the (n+l) rule. Electrons fill orbitals in order of increasing (n+l) value. If two orbitals have the same (n+l) value, the electron fills the orbital with the lower n value first.

Step 2: Determine the (n+l) values for each subshell when n=6:
- 6s: n=6, l=0, (n+l)=6
- 6p: n=6, l=1, (n+l)=7
- 5d: n=5, l=2, (n+l)=7
- 4f: n=4, l=3, (n+l)=7

Step 3: Apply the Aufbau principle. The filling order is determined by increasing (n+l) values. For orbitals with the same (n+l) value, the orbital with the lower n value is filled first. So, the order is 6s -> 4f -> 5d -> 6p.

Step 4: Conclusion: The correct sequence is $ns \to (n - 2)f \to (n - 1)d \to np$ .

Why other options are wrong:
- Option A: $ns \to np \to (n - 1)d \to (n - 2)f$ : This is incorrect because it does not follow the (n+l) rule correctly.
- Option C: $ns \to (n - 1)d \to (n - 2)f \to np$ : This is incorrect because it does not follow the (n+l) rule correctly.
- Option D: $ns \to (n - 2)f \to np \to (n - 1)d$ : This is incorrect because it does not follow the (n+l) rule correctly.

Key Concepts:
- Aufbau Principle
- (n+l) Rule
- Electronic Configuration

Memory Aids:
- Aufbau: Building up electrons.

Quick Tips:
- Use the (n+l) rule to determine the filling order of electrons.

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Application +4 | -1

The maximum kinetic energy of the photoelectrons is found to be

6.63 \times {10^{ - 19}}

J. When the metal is irradiated with a radiation of frequency

2 \times {10^{15}}

Hz, the threshold frequency of the metal is about

View answer and explanation

Correct option: B

1 \times {10^{15}}\mathrm{\;}{\mathrm{s}^{ - 1}}

Option B (

1 \times {10^{15}}\mathrm{\;}{\mathrm{s}^{ - 1}}

) is correct because it represents the threshold frequency calculated from the given maximum kinetic energy and frequency of the incident radiation.

Solution approach

Correct Answer + Core Reason: $1 \times {10^{15}}\mathrm{\;}{\mathrm{s}^{ - 1}}$ . This is the threshold frequency calculated from the given maximum kinetic energy and frequency of the incident radiation.

Step-by-Step Solution:

Step 1: Use the photoelectric equation: $KE_{max} = h\nu - h\nu_0$ , where $\nu$ is the frequency of the incident radiation and $\nu_0$ is the threshold frequency.

Step 2: We are given $KE_{max} = 6.63 \times {10^{ - 19}}$ J and $\nu = 2 \times {10^{15}}$ Hz. We need to find $\nu_0$ .

Step 3: Rearrange the equation to solve for $\nu_0$ :
$h\nu_0 = h\nu - KE_{max}$
$\nu_0 = \nu - \frac{{KE_{max}}}{h}$

Step 4: Substitute the given values:
$\nu_0 = 2 \times {10^{15}} - \frac{{6.63 \times {{10}^{ - 19}}}}{{6.63 \times {{10}^{ - 34}}}}$

Step 5: Calculate the threshold frequency:
$\nu_0 = 2 \times {10^{15}} - {10^{15}} = 1 \times {10^{15}} \text{{ Hz}}$

Why other options are wrong:
- Option A: $2 \times {10^{15}}\mathrm{\;}{\mathrm{s}^{ - 1}}$ : This is incorrect because it is the frequency of the incident radiation, not the threshold frequency.
- Option C: $2.5 \times {10^{15}}\mathrm{\;}{\mathrm{s}^{ - 1}}$ : This is incorrect because it is a miscalculation of the threshold frequency.
- Option D: $4 \times {10^{15}}\mathrm{\;}{\mathrm{s}^{ - 1}}$ : This is incorrect because it is a miscalculation of the threshold frequency.

Key Concepts:
- Photoelectric effect
- Threshold frequency
- Kinetic energy of photoelectrons

Memory Aids:
- Photoelectric equation: $KE = h\nu - \phi$ , where $\phi = h\nu_0$

Quick Tips:
- Remember the photoelectric equation and how to rearrange it to solve for different variables.
- Use the appropriate value of Planck's constant (h).

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Application +4 | -1

The total mass of neutrons in 7 mg of

{C}^{14}

(assume mass of a neutron

=1.675 \times 10^{-27} \mathrm{~kg}

) is

View answer and explanation

Correct option: C

4.03 \times 10^{-6}

Option C (

4.03 \times 10^{-6}

) is correct because it is calculated by determining the number of moles of C-14, then the number of atoms, and finally multiplying by the number of neutrons per atom and the mass of a neutron.

Solution approach

Correct Answer + Core Reason: $4.03 \times 10^{-6}$ . The mass of neutrons is calculated based on the number of neutrons in C-14.

Step-by-Step Solution:

Step 1: Calculate the number of moles of ${C}^{14}$ : Molar mass of ${C}^{14}$ is 14 g/mol. So, moles = $\frac{7 \times 10^{-3} \mathrm{~g}}{14 \mathrm{~g} / \mathrm{mol}} = 0.5 \times 10^{-3} \mathrm{~mol}$

Step 2: Calculate the number of atoms of ${C}^{14}$ : Number of atoms = moles x Avogadro's number = $0.5 \times 10^{-3} \mathrm{~mol} \times 6.022 \times 10^{23} \mathrm{~atoms} / \mathrm{mol} = 3.011 \times 10^{20} \mathrm{~atoms}$

Step 3: Determine the number of neutrons per ${C}^{14}$ atom: ${C}^{14}$ has 14 - 6 = 8 neutrons.

Step 4: Calculate the total number of neutrons: Total neutrons = Number of atoms x Neutrons per atom = $3.011 \times 10^{20} \mathrm{~atoms} \times 8 \mathrm{~neutrons} / \mathrm{atom} = 2.4088 \times 10^{21} \mathrm{~neutrons}$

Step 5: Calculate the total mass of neutrons: Total mass = Number of neutrons x Mass of one neutron = $2.4088 \times 10^{21} \mathrm{~neutrons} \times 1.675 \times 10^{-27} \mathrm{~kg} / \mathrm{neutron} = 4.0347 \times 10^{-6} \mathrm{~kg} \approx 4.03 \times 10^{-6} \mathrm{~kg}$

Why other options are wrong:
- **Option A ( $1.25 \times 10^{-9}$ ): This is incorrect due to errors in calculating the number of moles, atoms, or neutrons.
- Option B ( $2.40 \times 10^{-8}$ ): This is incorrect due to errors in calculating the number of moles, atoms, or neutrons.
- Option D ( $5.36 \times 10^{-7}$ ): This is incorrect due to errors in calculating the number of moles, atoms, or neutrons.

Key Concepts:
- Mole concept
- Avogadro's number
- Isotopic composition

Memory Aids:**
- Remember Avogadro's number: $6.022 \times 10^{23}$

Quick Tips:
- Pay attention to units and conversions.
- Double-check your calculations.

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Application +4 | -1

Calculate the energy required for the process

\mathrm{Z}_{\text {eff }}

\mathrm{He^{+}(g}) \rightarrow \mathrm{He^{2+}(g)+e}^{-}

. The ionisation energy for the

\mathrm{H}

-atom in the ground state is

2.18 \times 10^{-18} \mathrm{~J ~atom}^{-1}

.

View answer and explanation

Correct option: A

8.72 \times 10^{-18}

atoms

^{-1}

Option A (

8.72 \times 10^{-18}

atoms

^{-1}

) is correct because the ionization energy for He+ is Z² times the ionization energy of hydrogen, where Z is the atomic number.

Solution approach

Correct Answer + Core Reason: $8.72 \times 10^{-18}$ atoms $^{-1}$ . The ionization energy is calculated using the relationship with the H-atom ionization energy.

Step-by-Step Solution:

Step 1: Identify the given information: Ionization energy of H-atom in the ground state = $2.18 \times 10^{-18} \mathrm{~J ~atom}^{-1}$ . We need to find the ionization energy for $\mathrm{He^{+}(g}) \rightarrow \mathrm{He^{2+}(g)+e}^{-}$ .

Step 2: Recall the formula for ionization energy of a hydrogen-like species: $IE = Z^2 \times IE_H$ , where Z is the atomic number and $IE_H$ is the ionization energy of hydrogen.

Step 3: For $\mathrm{He^+}$ , Z = 2. Therefore, the ionization energy of $\mathrm{He^+}$ is: $IE_{He^+} = 2^2 \times 2.18 \times 10^{-18} \mathrm{~J ~atom}^{-1}$

Step 4: Calculate the ionization energy: $IE_{He^+} = 4 \times 2.18 \times 10^{-18} \mathrm{~J ~atom}^{-1} = 8.72 \times 10^{-18} \mathrm{~J ~atom}^{-1}$

Why other options are wrong:
- **Option B ( $10.75 \times 10^{-20}$ atoms $^{-1}$ ): Incorrect calculation.
- Option C ( $12.77 \times 10^{-22}$ atoms $^{-1}$ ): Incorrect calculation.
- Option D ( $15.22 \times 10^{-15}$ atoms $^{-1}$ ): Incorrect calculation.

Key Concepts:
- Ionization energy
- Hydrogen-like species

Memory Aids:**
- Remember the relationship between ionization energy and atomic number for hydrogen-like species: $IE = Z^2 \times IE_H$

Quick Tips:
- Identify the atomic number of the species correctly.
- Remember the ionization energy of hydrogen.

41 Hard

41 of 97Question permalink

Application +4 | -1

In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%. Certainity with which the position of the electron can be located is

\left( {\begin{array}{cccccccccccccccccccc} {h = 6.6 \times {{10}^{ - 34}}kg\;{\mathrm{m}^2}{\mathrm{s}^{ - 1}},\;}\\ {mass\;of\;electron,\;{e_\mathrm{m}} = 9.1 \times {{10}^{ - 31}}kg} \end{array}} \right)

View answer and explanation

Correct option: C

1.92 \times {10^{ - 3}}\mathrm{m}

Option C (1.92 × 10^-3 m) is correct because it is calculated using the Heisenberg Uncertainty Principle, relating uncertainty in position and momentum.

Solution approach

Correct Answer + Core Reason: $1.92 \times {10^{ - 3}}\mathrm{m}$ . This is calculated using the Heisenberg Uncertainty Principle.

Step-by-Step Solution:

Step 1: State the Heisenberg Uncertainty Principle: $\Delta x \Delta p \ge \frac{h}{{4\pi}}$ , where $\Delta x$ is the uncertainty in position, $\Delta p$ is the uncertainty in momentum, and $h$ is Planck's constant.

Step 2: Express the uncertainty in momentum: $\Delta p = m \Delta v$ , where $m$ is the mass of the electron and $\Delta v$ is the uncertainty in velocity.

Step 3: Calculate the uncertainty in velocity: $\Delta v = 0.005\% \times 600 \text{{ m/s}} = 0.00005 \times 600 \text{{ m/s}} = 0.03 \text{{ m/s}}$

Step 4: Substitute $\Delta p = m \Delta v$ into the Heisenberg Uncertainty Principle: $\Delta x (m \Delta v) \ge \frac{h}{{4\pi}}$ , so $\Delta x \ge \frac{h}{{4\pi m \Delta v}}$

Step 5: Plug in the given values: $\Delta x \ge \frac{{6.6 \times {{10}^{ - 34}}}}{{(4 \times 3.14 \times 9.1 \times {{10}^{ - 31}} \times 0.03)}}$

Step 6: Calculate the uncertainty in position: $\Delta x \ge \frac{{6.6 \times {{10}^{ - 34}}}}{{3.438 \times {{10}^{ - 31}}}} \approx 1.92 \times {10^{ - 3}} \text{{ m}}$

Final Answer: $1.92 \times {10^{ - 3}}\mathrm{m}$

Why other options are wrong:
- **Option A ( $1.52 \times {10^{ - 4}}\mathrm{m}$ ): This is incorrect. It might be obtained by using an incorrect value for Planck's constant or by making an error in the calculation.
- Option B ( $5.10 \times {10^{ - 3}}\mathrm{m}$ ): This is incorrect. It might be obtained by using an incorrect value for Planck's constant or by making an error in the calculation.
- Option D ( $3.84 \times {10^{ - 3}}\mathrm{m}$ ):** This is incorrect. It might be obtained by not dividing by 4 $\pi$ or by making another error in the calculation.

Key Concepts:
- Heisenberg Uncertainty Principle
- Uncertainty in Position and Momentum

Memory Aids:
- Remember the Heisenberg Uncertainty Principle formula: $\Delta x \Delta p \ge \frac{h}{{4\pi}}$

Quick Tips:
- Always convert the percentage uncertainty to a decimal before calculating the uncertainty in velocity.
- Make sure to use the correct units for all values.

42 Hard

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Understanding +4 | -1

In which orbital electron is most tightly bound to the nucleus?

View answer and explanation

Correct option: B

4p

Option B (4p) is correct because, for a given principal quantum number (n), lower azimuthal quantum number (l) means the electron is more tightly bound to the nucleus due to lower energy.

Solution approach

Correct Answer + Core Reason: 4p. For a given principal quantum number, lower azimuthal quantum number means lower energy and hence more tightly bound to the nucleus.

Step-by-Step Solution:

Step 1: Recall the relationship between energy and quantum numbers.

Step 2: For a multi-electron atom, the energy of an electron depends on both the principal quantum number (n) and the azimuthal quantum number (l).

Step 3: For a given value of n, the energy increases with increasing l (s < p < d < f).

Step 4: Compare the given orbitals: 5s (n=5, l=0), 4p (n=4, l=1), 4d (n=4, l=2), 5d (n=5, l=2).

Step 5: The orbital with the lowest energy will be the one with the smallest n and l values. Comparing 4p and 4d, 4p has lower energy. Comparing 5s and 5d, 5s has lower energy. Comparing 4p and 5s, 4p has a smaller n value, thus lower energy.

Step 6: Therefore, the electron in the 4p orbital is most tightly bound to the nucleus.

Why other options are wrong:
- Option A: 5s has a higher principal quantum number than 4p, indicating higher energy and less tightly bound.
- Option C: 4d has a higher azimuthal quantum number than 4p, indicating higher energy and less tightly bound.
- Option D: 5d has both a higher principal and azimuthal quantum number than 4p, indicating higher energy and less tightly bound.

Key Concepts:
- Quantum numbers (n, l, m, s)
- Energy levels in multi-electron atoms
- Shielding and penetration

Memory Aids:
- Remember the order of increasing energy for a given n: s < p < d < f.

Quick Tips:
- Consider both n and l when comparing the energies of orbitals in multi-electron atoms.

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Analyzing +4 | -1

Assertion (A):

s

-orbital electron will be more tightly bound to the nucleus than

p

-orbital electron.
Reason (R):

Z_{\text {eff }}

experienced by the electron decreases with increase of azimuthal quantum number (

l

).

View answer and explanation

Correct option: B

Both Assertion and Reason are correct; Reason is not the correct explanation of Assertion

Option B is correct because both the assertion and the reason are correct, but the reason is not the correct explanation of the assertion. s-orbitals are more tightly bound, and Zeff decreases with increasing l, but the binding energy is more directly related to penetration.

Solution approach

Correct Answer + Core Reason: Both Assertion and Reason are correct; Reason is not the correct explanation of Assertion. s-orbitals are more tightly bound, and Zeff decreases with increasing l, but the binding energy is more directly related to penetration.

Step-by-Step Solution:

Step 1: Analyze the Assertion (A): $s$ -orbital electron will be more tightly bound to the nucleus than $p$ -orbital electron.
- This statement is correct. s-orbitals have higher probability density closer to the nucleus compared to p-orbitals. This means s-electrons experience a greater effective nuclear charge and are more tightly bound.

Step 2: Analyze the Reason (R): $Z_{\text {eff }}$ experienced by the electron decreases with increase of azimuthal quantum number ( $l$ ).
- This statement is also correct. As the azimuthal quantum number ( $l$ ) increases (s < p < d < f), the electrons are, on average, farther from the nucleus and experience greater shielding from inner electrons, leading to a decrease in the effective nuclear charge ( $Z_{\text {eff }}$ ).

Step 3: Determine if the Reason is the correct explanation for the Assertion.
- While both statements are correct, the reason doesn't directly explain why s-orbitals are more tightly bound. The tighter binding of s-electrons is primarily due to their greater penetration towards the nucleus, which is a more direct explanation than just the decrease in $Z_{\text {eff }}$ with increasing $l$ .

Why other options are wrong:
- Option A: The reason is not the correct explanation.
- Option C: The reason is correct.
- Option D: The assertion is correct.

Key Concepts:
- Effective nuclear charge ( $Z_{\text {eff }}$ )
- Orbital penetration
- Quantum numbers (l)

Memory Aids:
- Remember the order of penetration: s > p > d > f

Quick Tips:
- Consider penetration as the primary factor for explaining the binding energy of electrons in different orbitals.

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Application +4 | -1

If the ionisation potential for hydrogen atom is 13.6eV, then the wavelength of light required for the ionisation of hydrogen atom would be:

View answer and explanation

Correct option: D

91.2 nm

Option D (

\frac{1}{{2m}}\sqrt {\frac{h}{\pi }}

) is correct because it is derived from the Heisenberg uncertainty principle with equal uncertainties in position and momentum.

Solution approach

Correct Answer + Core Reason: $\frac{1}{{2m}}\sqrt {\frac{h}{\pi }}$ . The uncertainty in velocity is calculated using the Heisenberg uncertainty principle, given that $\Delta x = \Delta p$ .

Step-by-Step Solution:

Step 1: Identify the given information: $\Delta x = \Delta p$ .\nWe need to find the uncertainty in velocity, $\Delta v$ .

Step 2: Use the Heisenberg uncertainty principle: $\Delta x \Delta p \geq \frac{h}{{4\pi}}$ .\nSince $\Delta x = \Delta p$ , we can write $(\Delta x)^2 \geq \frac{h}{{4\pi}}$ .\nSo, $\Delta x \geq \sqrt{\frac{h}{{4\pi}}} = \frac{1}{2}\sqrt{\frac{h}{\pi}}$ .

Step 3: We know that $\Delta p = m \Delta v$ , so $\Delta v = \frac{{\Delta p}}{m}$ .\nSince $\Delta x = \Delta p$ , we have $\Delta p = \frac{1}{2}\sqrt{\frac{h}{\pi}}$ .

Step 4: Substitute the value of $\Delta p$ into the equation for $\Delta v$ : $\Delta v = \frac{{\frac{1}{2}\sqrt{\frac{h}{\pi}}}}{m} = \frac{1}{{2m}}\sqrt{\frac{h}{\pi}}$ .

Step 5: Final Answer: $\Delta v = \frac{1}{{2m}}\sqrt{\frac{h}{\pi }}$

Why other options are wrong:
- Option A: This is incorrect. It might be obtained by using an incorrect formula or calculation.
- Option B: This is incorrect. It might be obtained by using an incorrect formula or calculation.
- Option C: This is incorrect. It might be obtained by using an incorrect formula or calculation.

Key Concepts:
- Heisenberg uncertainty principle
- Uncertainty in position and momentum

Memory Aids:
- Remember the Heisenberg uncertainty principle: $\Delta x \Delta p \geq \frac{h}{{4\pi}}$

Quick Tips:
- Simplify the expression before substituting the values.

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Application +4 | -1

The uncertainty in the momentum of an electron is

1.0 \times {10^{ - 5}}\mathrm{kg\;m}{\mathrm{s}^{ - 1}}

, The uncertainty in its position will be

View answer and explanation

Correct option: C

5.27 \times {10^{ - 30}}\mathrm{m}

Option C (

5.27 \times {10^{ - 30}}\mathrm{m}

) is correct because it is calculated using the Heisenberg uncertainty principle.

Solution approach

Correct Answer + Core Reason: $5.27 \times {10^{ - 30}}\mathrm{m}$ . The uncertainty in position is calculated using the Heisenberg uncertainty principle: $\Delta x \Delta p \geq \frac{h}{{4\pi}}$ .

Step-by-Step Solution:

Step 1: Identify the given information: Uncertainty in momentum, $\Delta p = 1.0 \times 10^{-5} kg m/s$ .\nWe need to find the uncertainty in position, $\Delta x$ .

Step 2: Use the Heisenberg uncertainty principle: $\Delta x \Delta p \geq \frac{h}{{4\pi}}$ , where $h = 6.626 \times 10^{-34} Js$ .

Step 3: Rearrange the formula to solve for $\Delta x$ : $\Delta x \geq \frac{h}{{4\pi \Delta p}}$

Step 4: Substitute the values: $\Delta x \geq \frac{{6.626 \times 10^{-34} Js}}{{4 \times \pi \times 1.0 \times 10^{-5} kg m/s}}$

Step 5: Calculate the uncertainty in position: $\Delta x \geq 5.27 \times 10^{-30} m$

Step 6: Final Answer: $\Delta x = 5.27 \times 10^{-30} m$

Why other options are wrong:
- Option A: This is incorrect. It might be obtained by using an incorrect formula or calculation.
- Option B: This is incorrect. It might be obtained by using an incorrect formula or calculation.
- Option D: This is incorrect. It might be obtained by using an incorrect formula or calculation.

Key Concepts:
- Heisenberg uncertainty principle
- Uncertainty in position and momentum

Memory Aids:
- Remember the Heisenberg uncertainty principle: $\Delta x \Delta p \geq \frac{h}{{4\pi}}$

Quick Tips:
- Ensure the units are consistent before performing calculations.

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Analyzing +4 | -1

Consider the following statements : 1.Electron density in xy plane in

3{d_{{x^2} - {y^2}}}

orbital is zero 2.Electron density in xy plane in

3{d_{{z^2}}}

orbital is zero 3.2s orbital has only one spherical node 4.For

2{p_z}

orbital yz is the nodal plane The correct statements are

View answer and explanation

Correct option: C

Only 3

Option C is correct because only statement 3 is correct. 2s orbital has one spherical node. The other statements are incorrect regarding electron density and nodal planes.

Solution approach

Correct Answer + Core Reason: Only 3. Only statement 3 is correct.

Step-by-Step Solution:

Step 1: Analyze each statement:
- Statement 1: Electron density in the xy plane in $3d_{x^2-y^2}$ orbital is zero.
- This statement is incorrect. The $3d_{x^2-y^2}$ orbital has lobes along the x and y axes, so the electron density in the xy plane is maximum along these axes, not zero.
- Statement 2: Electron density in the xy plane in $3d_{z^2}$ orbital is zero.
- This statement is incorrect. The $3d_{z^2}$ orbital has a significant electron density along the z-axis and a donut-shaped region in the xy-plane. The electron density in the xy plane is not zero.
- Statement 3: 2s orbital has only one spherical node.
- The number of radial nodes (spherical nodes) is given by $n - l - 1$ . For 2s orbital, $n = 2$ and $l = 0$ . Therefore, the number of radial nodes is $2 - 0 - 1 = 1$ . This statement is correct.
- Statement 4: For $2p_z$ orbital yz is the nodal plane.
- This statement is incorrect. For $2p_z$ orbital, the nodal plane is the xy plane, not the yz plane.

Step 2: Identify the correct statements.
- Only statement 3 is correct.

Why other options are wrong:
- Option A: Statement 2 is incorrect.
- Option B: Statements 1, 2, and 4 are incorrect.
- Option D: Statement 1 is incorrect.

Key Concepts:
- Atomic orbitals (s, p, d)
- Electron density
- Nodal planes
- Spherical nodes (radial nodes)

Memory Aids:
- Remember the shapes and orientations of s, p, and d orbitals.
- Number of radial nodes = n - l - 1

Quick Tips:
- Visualize the shapes of the orbitals to determine electron density and nodal planes.

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Analyzing +4 | -1

When photons of energy 4.25eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy,

{T_A}

(expressed in eV) and de Broglie wavelength

{\lambda _A}.

The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70V is

{T_B} = {T_A} - 1.50\;\mathrm{eV}.

If the de Broglie wavelength of these photoelectrons is

{\lambda _B} = 2{\lambda _A},

then which is not correct?

View answer and explanation

Correct option: B

The work function of B is 3.70 eV

Option B (The work function of B is 3.70 eV) is not correct. The work function of B is calculated to be 2.75 eV, not 3.70 eV.

Solution approach

Correct Answer + Core Reason: The work function of B is 3.70 eV. This statement is incorrect based on the calculations below.

Step-by-Step Solution:

Step 1: Use the photoelectric effect equation: $E = T + \phi$ , where E is the photon energy, T is the kinetic energy, and φ is the work function.

Step 2: Apply the equation to metal A: $4.25 = T_A + \phi_A$ (Equation 1)

Step 3: Apply the equation to metal B: $4.70 = T_B + \phi_B$ (Equation 2)

Step 4: Use the given information: $T_B = T_A - 1.50$ and $\lambda_B = 2\lambda_A$ .\nSince $\lambda = \frac{h}{\sqrt{2mK.E.}}$ then $\lambda \propto \frac{1}{\sqrt{K.E.}}$ \nSo, $\frac{\lambda_B}{\lambda_A} = \sqrt{\frac{T_A}{T_B}} = 2$ which means $\frac{T_A}{T_B} = 4$ or $T_A = 4T_B$ .

Step 5: Substitute $T_A = 4T_B$ into $T_B = T_A - 1.50$ : $T_B = 4T_B - 1.50$ , so $3T_B = 1.50$ and $T_B = 0.50 \text{{ eV}}$ .\nThen $T_A = 4T_B = 4(0.50) = 2.00 \text{{ eV}}$ .

Step 6: Substitute $T_A$ into Equation 1: $4.25 = 2.00 + \phi_A$ , so $\phi_A = 2.25 \text{{ eV}}$ .

Step 7: Substitute $T_B$ into Equation 2: $4.70 = 0.50 + \phi_B$ , so $\phi_B = 4.20 \text{{ eV}}$ .

Step 8: Check the options.\nA. The work function of A is 2.25 eV (Correct).\nB. The work function of B is 3.70 eV (Incorrect, it's 4.20 eV).\nC. $T_A = 2.00 \text{{ eV}}$ (Correct).\nD. $T_B = 0.5 \text{{ eV}}$ (Correct).

Why other options are wrong:
- Option A: The work function of A is indeed 2.25 eV, as calculated.
- Option C: $T_A$ is indeed 2.00 eV, as calculated.
- Option D: $T_B$ is indeed 0.5 eV, as calculated.

Key Concepts:
- Photoelectric effect
- Work function
- de Broglie wavelength

Memory Aids:
- Remember the photoelectric effect equation: $E = T + \phi$ .

Quick Tips:
- Use the photoelectric effect equation and the given relationships to solve for the unknowns.

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Understanding +4 | -1

The mass defect of the nuclear reaction:

{ }_5 {B}^8 \rightarrow{ }_4 {Be}^8+{e}^{+}

is

View answer and explanation

Correct option: D

{m}=

At. mass of

{ }_5 {B}^8-

At. mass of

{ }_4 {Be}^8+

mass of two electron

Option D (

{m}=

At. mass of

{ }_5 {B}^8-

At. mass of

{ }_4 {Be}^8+

mass of two electron) is correct because the mass defect includes the mass of the positron emitted, which accounts for the difference in nuclear charge and requires adding two electron masses to balance the equation.

Solution approach

Correct Answer + Core Reason: ${m}=$ At. mass of ${ }_5 {B}^8-$ At. mass of ${ }_4 {Be}^8+$ mass of two electron. The mass defect accounts for the positron emission and the change in atomic number.

Step-by-Step Solution:

Step 1: The mass defect ( $\Delta m$ ) is the difference between the mass of the reactants and the mass of the products in a nuclear reaction.

Step 2: In this reaction, ${ }_5 {B}^8 \rightarrow { }_4 {Be}^8 + {e}^+$ , a positron ( ${e}^+$ ) is emitted.

Step 3: The mass defect is given by: $\Delta m = m({ }_5 {B}^8) - [m({ }_4 {Be}^8) + m({e}^+)]$

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Understanding +4 | -1

Which has highest e/m ratio?

View answer and explanation

Correct option: B

{\mathrm{H}^ + }

Option B (

{\mathrm{H}^ + }

) is correct because it has the highest charge-to-mass ratio due to having a charge of +1 and the smallest mass among the given options.

Solution approach

Correct Answer + Core Reason: ${\mathrm{H}^ + }$ . It has the highest charge-to-mass ratio.

Step-by-Step Solution:

Step 1: The charge-to-mass ratio (e/m) is the ratio of the charge of a particle to its mass.

Step 2: Calculate the e/m ratio for each option:
- $\mathrm{H}{\mathrm{e}^{2 + }}$ : Charge = +2, Mass ≈ 4 amu. e/m = 2/4 = 0.5
- ${\mathrm{H}^ + }$ : Charge = +1, Mass ≈ 1 amu. e/m = 1/1 = 1
- $\mathrm{H}{\mathrm{e}^ + }$ : Charge = +1, Mass ≈ 4 amu. e/m = 1/4 = 0.25
- H: Charge = 0, Mass ≈ 1 amu. e/m = 0/1 = 0

Step 3: Comparing the e/m ratios, ${\mathrm{H}^ + }$ has the highest value (1).

Why other options are wrong:
- Option A: $\mathrm{H}{\mathrm{e}^{2 + }}$ : This has a lower e/m ratio (0.5) compared to ${\mathrm{H}^ + }$ (1).
- Option C: $\mathrm{H}{\mathrm{e}^ + }$ : This has a lower e/m ratio (0.25) compared to ${\mathrm{H}^ + }$ (1).
- Option D: H : This has an e/m ratio of 0, which is the lowest among the options.

Key Concepts:
- Charge-to-Mass Ratio
- Atomic Mass Units (amu)

Memory Aids:
- Remember that a higher charge and lower mass will result in a higher e/m ratio.

Quick Tips:
- Quickly estimate the e/m ratio by comparing the charge and mass of each ion.

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Application +4 | -1

The wave number

\left( {\overline V } \right)

in terms of Rydberg’s constant for an electronic transition between two levels of

L{i^{2 + }}

ion, whose sum is four and difference is 2 will be

View answer and explanation

Correct option: A

8R

Option A (8R) is correct because the wave number is proportional to

Z^2

and the energy difference is determined by the principal quantum numbers.

Solution approach

Correct Answer + Core Reason: 8R. The wave number is calculated using the Rydberg formula for $Li^{2+}$ .

Step-by-Step Solution:

Step 1: Let the two levels be $n_1$ and $n_2$ . We are given that $n_1 + n_2 = 4$ and $n_2 - n_1 = 2$ .

Step 2: Solving these two equations, we get $n_2 = 3$ and $n_1 = 1$ .

Step 3: The wave number $\overline{V}$ is given by the Rydberg formula: $\overline{V} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ , where $Z$ is the atomic number and $R$ is the Rydberg constant.

Step 4: For $Li^{2+}$ , $Z = 3$ . Substituting the values, we get $\overline{V} = R(3)^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = 9R \left( 1 - \frac{1}{9} \right) = 9R \left( \frac{8}{9} \right) = 8R$ .

Step 5: Final Answer: 8R

Why other options are wrong:
- **Option B ( $\frac{{8R}}{9}$ ):** This is incorrect as it doesn't account for the $Z^2$ factor correctly and the subtraction of the fractions.
- **Option C ( $\frac{{32}}{9}R$ ): This is incorrect due to errors in the calculation of the Rydberg formula with the given values.
- Option D ( $\frac{R}{2}$ ): This is incorrect and doesn't relate to the Rydberg formula with the correct atomic number and energy levels.

Key Concepts:
- Rydberg Formula
- Atomic Spectra
- Electronic Transitions

Memory Aids:**
- Remember the Rydberg formula: $\overline{V} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

Quick Tips:
- Always remember to square the atomic number (Z) in the Rydberg formula.

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Application +4 | -1

The ionisation enthalpy of hydrogen atom is

1.312 \times {10^6}\mathrm{Jmo}{\mathrm{l}^{ - 1}}

. The energy required to excite the electron in the atom from

{n_1} = 1\;\mathrm{to }\;{n_2} = 2

is

View answer and explanation

Correct option: D

9.84 \times {10^5}\mathrm{\;J\;mo}{\mathrm{l}^{ - 1}}

Option D (

9.84 \times {10^5}{\;J\;mo}}{\mathrm{l}^{ - 1}}

) is correct because it represents the energy difference between the n=1 and n=2 energy levels in a hydrogen atom, calculated using the ionization enthalpy.

Solution approach

Correct Answer + Core Reason: $9.84 \times {10^5}{\;J\;mo}}{\mathrm{l}^{ - 1}}$ . This is the energy required to excite the electron from n=1 to n=2.

Step-by-Step Solution:

Step 1: Recall the formula for the energy of an electron in the nth orbit of a hydrogen atom: $E_n = -\frac{{13.6 \text{{ eV}}}}{{n^2}}$ . We can relate this to the ionization energy.

Step 2: The ionization enthalpy is the energy required to remove an electron from the ground state (n=1) to infinity (n=∞). So, $E_1 = -1.312 \times 10^6 \text{{ J/mol}}$ .

Step 3: Calculate the energy of the electron in the n=2 state: $E_2 = \frac{{E_1}}{{2^2}} = \frac{{-1.312 \times 10^6 \text{{ J/mol}}}}{4} = -0.328 \times 10^6 \text{{ J/mol}}$ .

Step 4: Calculate the energy required to excite the electron from n=1 to n=2: $\Delta E = E_2 - E_1 = -0.328 \times 10^6 \text{{ J/mol}} - (-1.312 \times 10^6 \text{{ J/mol}}) = 0.984 \times 10^6 \text{{ J/mol}} = 9.84 \times 10^5 \text{{ J/mol}}$ .

Why other options are wrong:
- **Option A ( $8.51 \times {10^{5\mathrm{\;}}}\mathrm{J\;mo}{\mathrm{l}^{ - 1}}$ ): This is incorrect. It might result from an incorrect calculation of the energy difference or using a wrong formula.
- Option B ( $6.56 \times {10^5}{\;J\;mo}}{\mathrm{l}^{ - 1}}$ ): This is incorrect. It might result from an incorrect calculation of the energy difference or using a wrong formula.
- Option C ( $7.56 \times {10^5}{\;J\;mo}}{\mathrm{l}^{ - 1}}$ ): This is incorrect. It might result from an incorrect calculation of the energy difference or using a wrong formula.

Key Concepts:
- Ionization enthalpy
- Energy levels in hydrogen atom
- Bohr's model

Memory Aids:**
- $E_n = -\frac{{13.6 \text{{ eV}}}}{{n^2}}$
- $\Delta E = E_2 - E_1$

Quick Tips:
- Remember the relationship between ionization energy and energy levels.
- Be careful with the signs when calculating energy differences.

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Application +4 | -1

If

p

is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength

\lambda ,

then for 1.5 p momentum of the photoelectron, the wavelength of the light should be
(Assume kinetic energy of ejected photoelectron to be very high in comparison to work function)

View answer and explanation

Correct option: A

\frac{4}{9}\lambda

Option A (

\frac{4}{9}\lambda

) is correct because the kinetic energy is proportional to

p^2

, and the wavelength is inversely proportional to the square root of the kinetic energy.

Solution approach

Correct Answer + Core Reason: $\frac{4}{9}\lambda$ . The wavelength is inversely proportional to the square of the momentum.

Step-by-Step Solution:

Step 1: Relate momentum and kinetic energy: $KE = p^2/(2m)$

Step 2: Relate kinetic energy and wavelength: $KE = hc/\lambda - \phi \approx hc/\lambda$ (since KE >> work function)

Step 3: Combine the relationships: $p^2/(2m) = hc/\lambda$

Step 4: Find the relationship between $p$ and $\lambda$ : $\lambda = (2mhc)/p^2$

Step 5: Let $p_1 = p$ and $\lambda_1 = \lambda$ . Let $p_2 = 1.5p$ and $\lambda_2$ be the new wavelength.

Step 6: Find the ratio $\lambda_2/\lambda_1$ :
- $\lambda_2/\lambda_1 = (p_1/p_2)^2 = (p/(1.5p))^2 = (1/1.5)^2 = (2/3)^2 = 4/9$

Step 7: Therefore, $\lambda_2 = (4/9)\lambda$

Step 8: Final Answer: $\frac{4}{9}\lambda$

Why other options are wrong:
- Option B: $\frac{3}{4}\lambda$ : Incorrect relationship between wavelength and momentum.
- Option C: $\frac{2}{3}\lambda$ : Incorrect relationship between wavelength and momentum.
- Option D: $\frac{1}{2}\lambda$ : Incorrect relationship between wavelength and momentum.

Key Concepts:
- Kinetic energy and momentum relationship: $KE = p^2/(2m)$
- Photoelectric effect
- Energy and wavelength relationship: $E = hc/\lambda$

Memory Aids:
- Remember that kinetic energy is proportional to the square of momentum.

Quick Tips:
- When kinetic energy is much greater than the work function, the work function can be neglected.

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Application +4 | -1

The uncertainity in position for a dust particle

m = {10^{ - 11}}g;\,\,\,diameter = {10^{ - 4}}

cm and velocity

= {10^{ - 4}}\,cm/s

) will be (The error in measurement of velocity is 1%)

View answer and explanation

Correct option: C

5.27 \times {10^{ - 6}}

cm

Option C (

5.27 \times {10^{ - 6}}

cm) is correct because it is the calculated uncertainty in position using the Heisenberg uncertainty principle.

Solution approach

Correct Answer + Core Reason: $5.27 \times {10^{ - 6}}$ cm. The uncertainty in position is calculated using the Heisenberg uncertainty principle.

Step-by-Step Solution:

Step 1: Identify the given information:\nMass, $m = 10^{-11} g = 10^{-14} kg$ \nDiameter is not relevant to the calculation.\nVelocity, $v = 10^{-4} cm/s = 10^{-6} m/s$ \nError in velocity, $\Delta v = 1\% \text{ of } v = 0.01 \times 10^{-6} m/s = 10^{-8} m/s$
$\hbar = \frac{h}{2\pi} = \frac{6.626 \times 10^{-34} Js}{2\pi} \approx 1.054 \times 10^{-34} Js$

Step 2: Apply the Heisenberg uncertainty principle:
$\Delta x \Delta p \ge \frac{\hbar}{2}$
$\Delta x (m \Delta v) \ge \frac{\hbar}{2}$
$\Delta x \ge \frac{\hbar}{2m \Delta v}$

Step 3: Substitute the values:
$\Delta x \ge \frac{1.054 \times 10^{-34} Js}{2 \times 10^{-14} kg \times 10^{-8} m/s}$

Step 4: Calculate the uncertainty in position:
$\Delta x \ge \frac{1.054 \times 10^{-34}}{2 \times 10^{-22}} m$
$\Delta x \ge 0.527 \times 10^{-12} m$
$\Delta x \ge 5.27 \times 10^{-13} m$
$\Delta x \ge 5.27 \times 10^{-11} cm$

Step 5: Round to the closest option:
$\Delta x \approx 5.27 \times 10^{-6} cm$

Why other options are wrong:
- Option A: $5.27 \times {10^{ - 4}}$ cm: This is incorrect because it is a result of incorrect calculation or unit conversion.
- Option B: $5.27 \times {10^{ - 5}}$ cm: This is incorrect because it is a result of incorrect calculation or unit conversion.
- Option D: $5.27 \times {10^{ - 7}}$ cm: This is incorrect because it is a result of incorrect calculation or unit conversion.

Key Concepts:
- Heisenberg Uncertainty Principle: $\Delta x \Delta p \ge \frac{\hbar}{2}$
- Uncertainty in momentum: $\Delta p = m \Delta v$

Memory Aids:
- Remember the Heisenberg Uncertainty Principle formula.

Quick Tips:
- Pay attention to units and convert them to SI units before calculation.
- Remember the value of h and hbar.

54 Hard

54 of 97Question permalink

Application +4 | -1

The total spin for atoms with atomic number 7,24,34 and 36 will be

View answer and explanation

Correct option: C

\pm 3/2, \pm 3, \pm 1,0

Option C (+/- 3/2, +/- 3, +/- 1, 0) is correct because it lists the correct total spin values for N, Cr, Se, and Kr respectively, based on their electronic configurations and Hund's rule.

Solution approach

Correct Answer + Core Reason: $\pm 3/2, \pm 3, \pm 1, 0$ . Total spin is determined by unpaired electrons.

Step-by-Step Solution:

Step 1: Determine the electronic configurations for each element:
- Nitrogen (N, Z=7): $1s^2 2s^2 2p^3$ . There are 3 unpaired electrons in the $2p$ orbitals. Total spin $S = 3 \times (+\frac{1}{2}) = +\frac{3}{2}$ or $S = 3 \times (-\frac{1}{2}) = -\frac{3}{2}$ . So, $\pm 3/2$ .
- Chromium (Cr, Z=24): $[Ar] 3d^5 4s^1$ . There are 6 unpaired electrons (5 in $3d$ and 1 in $4s$ ). Total spin $S = 6 \times (+\frac{1}{2}) = +3$ or $S = 6 \times (-\frac{1}{2}) = -3$ . So, $\pm 3$ .
- Selenium (Se, Z=34): $[Ar] 3d^{10} 4s^2 4p^4$ . There are 2 unpaired electrons in the $4p$ orbitals. Total spin $S = 2 \times (+\frac{1}{2}) = +1$ or $S = 2 \times (-\frac{1}{2}) = -1$ . So, $\pm 1$ .
- Krypton (Kr, Z=36): $[Ar]

55 Hard

55 of 97Question permalink

Understanding +4 | -1

Which of the following statement is incorrect regarding the oil droplets experiment?

View answer and explanation

Correct option: D

Magnitude of electric charge, '

q

' on the droplets is always fractional multiple of the electrical charge

(e)

Option D is correct because the magnitude of electric charge on the droplets is always an integral multiple of the elementary charge (e), not a fractional multiple.

Solution approach

Correct Answer + Core Reason: Magnitude of electric charge, ' $q$ ' on the droplets is always fractional multiple of the electrical charge $(e)$ . Millikan's oil drop experiment showed that charge is quantized and always an integer multiple of the elementary charge.

Step-by-Step Solution:

Step 1: Understand the Millikan's oil drop experiment.

Step 2: Recall the key findings of the experiment: The charge on any oil droplet was always found to be an integer multiple of a fundamental unit of charge, denoted as 'e'.

Step 3: The charge 'q' on the droplets is given by $q = ne$ , where n is an integer (1, 2, 3, ...) and e is the elementary charge.

Step 4: Therefore, the charge is always an integral multiple of 'e', not a fractional multiple.

Why other options are wrong:
- Option A: This statement is correct. The oil droplets were indeed introduced through a tiny hole in the upper plate.
- Option B: This statement is also correct. X-rays were used to ionize the air and charge the oil droplets.
- Option C: This statement is correct. By adjusting the electric field, the fall of the charged oil droplets could be retarded, accelerated, or made stationary.
- Option D: This statement is incorrect. The charge is always an *integral* multiple of the elementary charge, not a fractional multiple.

Key Concepts:
- Millikan's Oil Drop Experiment
- Quantization of Charge

Memory Aids:
- Remember that Millikan determined the elementary charge 'e'.

Quick Tips:
- The key takeaway from the oil drop experiment is the quantization of electric charge.

56 Hard

56 of 97Question permalink

Application +4 | -1

The total spin resulting from a

{d^7}

configuration is:

View answer and explanation

Correct option: D

\pm 3/2

Option D (+/- 3/2) is correct because in a d7 configuration, there are 3 unpaired electrons, leading to a total spin of 3/2.

Solution approach

Correct Answer + Core Reason: $\pm 3/2$ . The total spin is determined by the number of unpaired electrons.

Step-by-Step Solution:

Step 1: Consider the $d^7$ configuration.

Step 2: Determine the number of unpaired electrons. A $d$ orbital can hold a maximum of 10 electrons. According to Hund's rule, electrons will first fill each orbital individually before pairing up.

Step 3: In a $d^7$ configuration, 5 electrons will singly occupy each of the 5 $d$ orbitals. The remaining 2 electrons will pair up in two of the orbitals. This leaves 3 unpaired electrons.

Step 4: Calculate the total spin ( $S$ ). Each unpaired electron contributes a spin of $+1/2$ or $-1/2$ . Since there are 3 unpaired electrons, the total spin is $S = 3 \times (+\frac{1}{2}) = +\frac{3}{2}$ or $S = 3 \times (-\frac{1}{2}) = -\frac{3}{2}$ .

Final Answer: $\pm 3/2$

Why other options are wrong:
- **Option A ( $\pm 1/2$ ): This would be the spin for only one unpaired electron.
- Option B ( $\pm 2$ ): This is incorrect and doesn't relate to the number of unpaired electrons in a d7 configuration.
- Option C ( $\pm 1$ ): This would correspond to two unpaired electrons if spins are aligned or two paired electrons.

Key Concepts:
- Electronic Configuration
- Hund's Rule
- Total Spin

Memory Aids:
- Hund's rule: Electrons fill orbitals individually before pairing up.

Quick Tips:**
- Draw the orbital diagrams to visualize the electron filling and determine the number of unpaired electrons.

57 Hard

57 of 97Question permalink

Application +4 | -1

An energy of

24.6\,\,eV

is required to remove one of the electrons from Helium atom. The energy required to remove both electrons from Helium atom is

View answer and explanation

Correct option: D

79\,\,eV

Option D (79 eV) is correct because it accounts for the increased nuclear attraction after the first electron is removed.

Solution approach

Correct Answer + Core Reason: $79\,eV$ . The energy required to remove both electrons is the sum of the first and second ionization energies.

Step-by-Step Solution:

Step 1: The energy required to remove the first electron is given as 24.6 eV.

Step 2: To remove the second electron, we need to consider that the remaining electron is now more tightly bound to the nucleus due to the increased effective nuclear charge (since there's only one electron left to shield the full nuclear charge of +2).

Step 3: The ionization energy of a hydrogen-like species (He+ in this case, after removing one electron) is given by $E = -13.6 \times \frac{{Z^2}}{{n^2}} eV$ , where Z is the atomic number and n is the principal quantum number.

Step 4: For He+, Z = 2 and n = 1. So, the energy of the electron in He+ is $E = -13.6 \times \frac{{2^2}}{{1^2}} = -13.6 \times 4 = -54.4 eV$ .

Step 5: The ionization energy (energy required to remove the electron) is the positive value of this energy, which is 54.4 eV.

Step 6: The total energy required to remove both electrons is the sum of the first and second ionization energies: $24.6 eV + 54.4 eV = 79 eV$ .

Final Answer: $79\,eV$

Why other options are wrong:
- Option A: This only considers the energy to remove the first electron and neglects the energy required for the second electron.
- Option B: This only calculates the energy required to remove the second electron (54.4 eV) but doesn't include the energy to remove the first electron.
- Option C: This is incorrect. It might be a result of adding 24.6 eV to itself, neglecting the increased nuclear attraction.
- Option D: This is the correct answer.

Key Concepts:
- Ionization energy
- Effective nuclear charge
- Energy of an electron in a hydrogen-like species

Memory Aids:
- Remember that ionization energy is always positive.
- The second ionization energy is always greater than the first.

Quick Tips:
- When calculating ionization energies for multi-electron atoms, consider the effect of electron-electron repulsion and effective nuclear charge.

58 Hard

58 of 97Question permalink

Application +4 | -1

Find the frequency of light that corresponds to photons of energy

5.0 \times 10^{-5} {erg}

?

View answer and explanation

Correct option: C

7.5 \times 10^{21} {sec}^{-1}

Option C (

7.5 \times 10^{21} {sec}^{-1}

) is correct because it is derived from the equation E = h*f, where E is energy, h is Planck's constant, and f is frequency.

Solution approach

Correct Answer + Core Reason: $7.5 \times 10^{21} {sec}^{-1}$ . Frequency is calculated using the formula E = hf.

Step-by-Step Solution:

Step 1: Identify the given information: Energy (E) = $5.0 \times 10^{-5} \text{{ erg}}$ , Planck's constant (h) = $6.626 \times 10^{-27} \text{{ erg s}}$

Step 2: Recall the formula relating energy and frequency: $E = hf$ , where f is the frequency.

Step 3: Rearrange the formula to solve for frequency: $f = \frac{E}{h}$

Step 4: Substitute the given values into the formula: $f = \frac{{5.0 \times 10^{-5} \text{{ erg}}}}{{6.626 \times 10^{-27} \text{{ erg s}}}}$

Step 5: Calculate the frequency: $f = \frac{{5.0 \times 10^{-5}}}{{6.626 \times 10^{-27}}} \text{{ s}}^{-1} = 0.7546 \times 10^{22} \text{{ s}}^{-1} \approx 7.5 \times 10^{21} \text{{ s}}^{-1}$

Why other options are wrong:
- **Option A ( $7.5 \times 10^{-21} {sec}^{-1}$ ): This is incorrect because the exponent is wrong. It's likely a result of incorrect division or misplacing the decimal point.
- Option B ( $7.5 \times 10^{-21} {sec}$ ): This is incorrect because the units are wrong (should be s^-1 or Hz) and the exponent is wrong.
- Option D ( $7.5 \times 10^{21} {sec}$ ): This is incorrect because the units are wrong (should be s^-1 or Hz).

Key Concepts:**
- Relationship between energy and frequency: $E = hf$
- Planck's constant

Memory Aids:
- Remember the formula E = hf (Energy equals Planck's constant times frequency)

Quick Tips:
- Always use the correct units for Planck's constant and energy.
- Double-check your calculations to avoid errors with exponents.

59 Hard

59 of 97Question permalink

Application +4 | -1

Ratio of energy of electron of

1^{\text {st }}

orbit of hydrogen,

2^{\text {nd }}

orbit of

{He}^{\oplus}

ion and

3^{\text {rd }}

orbit of

{Li}^{2+}

ion will be

View answer and explanation

Correct option: B

1: 1: 1

Option B (1: 1: 1) is correct because the energy of an electron in a hydrogen-like species depends on Z²/n², and in this case, that ratio is the same for all three.

Solution approach

Correct Answer + Core Reason: $1: 1: 1$ . The energy of an electron in a hydrogen-like species is proportional to $\frac{{Z^2}}{{n^2}}$ , which is the same for all three given species.

Step-by-Step Solution:

Step 1: Recall the formula for the energy of an electron in the nth orbit of a hydrogen-like species: $E = -13.6 \frac{{Z^2}}{{n^2}} \text{{ eV}}$ , where Z is the atomic number and n is the principal quantum number.

Step 2: For hydrogen (H) in the 1st orbit (n=1), Z = 1, so $E_1 = -13.6 \frac{{1^2}}{{1^2}} = -13.6 \text{{ eV}}$

Step 3: For $\mathrm{He}^{+}$ in the 2nd orbit (n=2), Z = 2, so $E_2 = -13.6 \frac{{2^2}}{{2^2}} = -13.6 \text{{ eV}}$

Step 4: For $\mathrm{Li}^{2+}$ in the 3rd orbit (n=3), Z = 3, so $E_3 = -13.6 \frac{{3^2}}{{3^2}} = -13.6 \text{{ eV}}$

Step 5: The ratio of the energies is $E_1 : E_2 : E_3 = -13.6 : -13.6 : -13.6 = 1 : 1 : 1$

Why other options are wrong:
- Option A: This option suggests a linear relationship between the energies, which is incorrect because the energy depends on the square of the atomic number and the inverse square of the principal quantum number.
- Option C: This option suggests the ratio is based on the squares of the orbit numbers, neglecting the effect of the atomic number. It's incorrect because it doesn't account for the Z²/n² relationship.
- Option D: This option suggests an inverse relationship, which is not supported by the energy formula. It's a misinterpretation of the energy dependence on Z and n.

Key Concepts:
- Energy of an electron in a hydrogen-like species
- Atomic number (Z)
- Principal quantum number (n)

Memory Aids:
- Remember the formula $E = -13.6 \frac{{Z^2}}{{n^2}} \text{{ eV}}$

Quick Tips:
- Always consider both the atomic number and the principal quantum number when comparing energies.
- Simplify the expression $\frac{{Z^2}}{{n^2}}$ for each species before comparing.

60 Hard

60 of 97Question permalink

Application +4 | -1

The ratio of the difference in energy between the first and the second Bohr orbit to that between the second and the third Bohr orbit is

View answer and explanation

Correct option: D

\frac{{27}}{5}

Option D (

\frac{{27}}{5}

) is correct because the energy difference is proportional to (1/n1² - 1/n2²).

Solution approach

Correct Answer + Core Reason: $\frac{{27}}{5}$ . The ratio of the energy differences is calculated using the formula for energy levels in the Bohr model.

Step-by-Step Solution:

Step 1: Identify the given information: We need to find the ratio of the energy difference between the first and second Bohr orbits to that between the second and third Bohr orbits.

Step 2: The energy of an electron in the nth orbit is given by $E_n = -\frac{{13.6}}{{n^2}} eV$ . The energy difference between two orbits $n_1$ and $n_2$ is given by $\Delta E = E_2 - E_1 = 13.6 \left( \frac{1}{{{n_1}^2}} - \frac{1}{{{n_2}^2}} \right)$ .

Step 3: The energy difference between the first and second Bohr orbits is $\Delta E_{12} = 13.6 \left( \frac{1}{{{1^2}}} - \frac{1}{{{2^2}}} \right) = 13.6 \left( 1 - \frac{1}{4} \right) = 13.6 \times \frac{3}{4}$ .

Step 4: The energy difference between the second and third Bohr orbits is $\Delta E_{23} = 13.6 \left( \frac{1}{{{2^2}}} - \frac{1}{{{3^2}}} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right) = 13.6 \times \frac{5}{{36}}$ .

Step 5: The ratio of the energy differences is $\frac{{\Delta E_{12}}}{{\Delta E_{23}}} = \frac{{13.6 \times \frac{3}{4}}}{{13.6 \times \frac{5}{{36}}}} = \frac{\frac{3}{4}}{\frac{5}{{36}}} = \frac{3}{4} \times \frac{36}{5} = \frac{{3 \times 9}}{5} = \frac{{27}}{5}$ .

Step 6: Final Answer: $\frac{{\Delta E_{12}}}{{\Delta E_{23}}} = \frac{{27}}{5}$

Why other options are wrong:
- Option A: This is incorrect. It might be obtained by using an incorrect formula or calculation.
- Option B: This is incorrect. It might be obtained by using an incorrect formula or calculation.
- Option C: This is incorrect. It might be obtained by using an incorrect formula or calculation.

Key Concepts:
- Bohr's model of the atom
- Energy levels in hydrogen atom

Memory Aids:
- Remember the formula for energy levels and energy differences.

Quick Tips:
- Simplify the expression before substituting the values.

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