A particle moving with a velocity of $\frac{1}{{100}}{\rm{th}}$ of that of light will cross a nucleus on about

$6\,\, \times \,\,{10^{ - 15}}\,\,{\rm{s}}$

ATOMS MCQs for NEET UG PHYSICS — Mixed difficulty

1 Medium

Analyzing +4 | -1

Which of the following statements are true regarding Bohr’s model of hydrogen atom

(I) Orbiting speed of electron decreases as it shifts to discrete orbits away from the nucleus

(II) Radii of allowed orbits of electron are proportional to the principal quantum number

(III) Frequency with which electrons orbit around the nucleus is discrete orbits is inversely proportional to the cube of principal quantum number

(IV) Binding force with which the electron is bound to the nucleus increases as it shifts to outer orbits

Select correct answer using the codes given below

View answer and explanation

Correct option: A

I and III

Option A (I and III) is correct because the orbiting speed decreases as the electron moves away from the nucleus, and the frequency is inversely proportional to the cube of the principal quantum number.

Solution approach

Correct Answer + Core Reason: I and III. Orbiting speed decreases with increasing orbit radius, and frequency is inversely proportional to the cube of the principal quantum number.

Step-by-Step Solution:

Step 1: Analyze statement I: Orbiting speed of electron decreases as it shifts to discrete orbits away from the nucleus.\nThe velocity of an electron in the nth orbit is given by $v = \frac{2.18 \times 10^6}{n} m/s$ . As $n$ increases (i.e., the electron shifts to orbits away from the nucleus), the velocity $v$ decreases. Therefore, statement I is true.

Step 2: Analyze statement II: Radii of allowed orbits of electron are proportional to the principal quantum number.\nThe radius of the nth orbit is given by $r = 0.529 n^2 \mathop A\limits^\circ$ . Therefore, the radius is proportional to $n^2$ , not $n$ . Statement II is false.

Step 3: Analyze statement III: Frequency with which electrons orbit around the nucleus is discrete orbits is inversely proportional to the cube of principal quantum number.\nThe frequency is given by $f = \frac{v}{2\pi r} = \frac{2.18 \times 10^6 / n}{2\pi (0.529 n^2 \times 10^{-10})} \propto \frac{1}{n^3}$ . Therefore, the frequency is inversely proportional to the cube of the principal quantum number. Statement III is true.

Step 4: Analyze statement IV: Binding force with which the electron is bound to the nucleus increases as it shifts to outer orbits.\nThe binding energy is given by $E = -\frac{13.6}{n^2} eV$ . As $n$ increases, the magnitude of the binding energy decreases, meaning the binding force decreases. Statement IV is false.

Step 5: Therefore, statements I and III are true.

Why other options are wrong:
- Option B: This is incorrect because statements II and IV are false.
- Option C: This is incorrect because statement II is false.
- Option D: This is incorrect because statements II and IV are false.

Key Concepts:
- Bohr's model of hydrogen atom
- Velocity of electron in nth orbit
- Radius of nth orbit
- Frequency of electron in nth orbit
- Binding energy

Memory Aids:
- Remember the formulas for velocity, radius, and energy in Bohr's model.

Quick Tips:
- Always remember the dependencies of velocity, radius, and energy on the principal quantum number.

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Application +4 | -1

A particle moving with a velocity of

\frac{1}{{100}}\mathrm{th}

of that of light will cross a nucleus on about

View answer and explanation

Correct option: D

{10^{ - 20}}\,\,\mathrm{s}

Option D (

{10^{ - 20}}\,\,\mathrm{s}

) is correct because it estimates the time it takes for a particle moving at 1/100th the speed of light to cross a nucleus, using the approximate size of a nucleus.

Solution approach

Correct Answer + Core Reason: ${10^{ - 20}}\,\,\mathrm{s}$ . This is the approximate time to cross a nucleus at 1/100th the speed of light.

Step-by-Step Solution:

Step 1: Estimate the size of the nucleus: The diameter of a nucleus is approximately $10^{-14}$ m.

Step 2: Calculate the velocity of the particle: $v = \frac{1}{100} c = \frac{1}{100} \times 3 \times 10^8 m/s = 3 \times 10^6 m/s$ .

Step 3: Calculate the time to cross the nucleus: $time = \frac{distance}{velocity} = \frac{10^{-14} m}{3 \times 10^6 m/s} \approx 0.33 \times 10^{-20} s \approx 10^{-20} s$ .

Why other options are wrong:
- **Option A ( ${10^{ - 8}}\,\,\mathrm{s}$ ): This is too large. It's likely a calculation error with the velocity or distance.
- Option B ( ${10^{ - 12}}\,\,\mathrm{s}$ ): This is also too large. It might be the time to cross an atom, not a nucleus.
- Option C ( $6\,\, \times \,\,{10^{ - 15}}\,\,\mathrm{s}$ ): This is also too large. It may result from using an incorrect nuclear size or velocity.

Key Concepts:
- Nuclear size
- Velocity
- Time calculation

Memory Aids:**
- Remember the approximate size of the nucleus: $10^{-14}$ m.

Quick Tips:
- Use order-of-magnitude estimations to quickly check your answer.

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Application +4 | -1

A hydrogen atom is in ground state absorbs a photon of ultraviolet radiation of wavelength

50 \mathrm{~nm}

. Assuming that the entire photon energy is taken up by electron with what KE will the electron be emitted.

View answer and explanation

Correct option: C

11.2 \mathrm{~eV}

Option C (11.2 eV) is correct because the kinetic energy of the emitted electron is the energy of the photon minus the ionization energy of hydrogen (13.6 eV).

Solution approach

Correct Answer + Core Reason: 11.2 eV. The kinetic energy is the photon energy minus the ionization energy of hydrogen.

Step-by-Step Solution:

Step 1: Calculate the energy of the photon using the formula: $E = \frac{hc}{\lambda}$ , where h is Planck's constant, c is the speed of light, and $\lambda$ is the wavelength.

Step 2: Convert the wavelength to meters: $\lambda = 50 \text{{ nm}} = 50 \times 10^{-9} \text{{ m}}$ .

Step 3: Calculate the photon energy in Joules: $E = \frac{(6.626 \times 10^{-34} \text{{ J s}})(3 \times 10^8 \text{{ m/s}})}{50 \times 10^{-9} \text{{ m}}} = 3.9756 \times 10^{-18} \text{{ J}}$ .

Step 4: Convert the photon energy to electron volts: $E = \frac{3.9756 \times 10^{-18} \text{{ J}}}{1.602 \times 10^{-19} \text{{ J/eV}}} = 24.816 \text{{ eV}}$ .

Step 5: Calculate the kinetic energy of the emitted electron: $KE = E_{photon} - IE$ , where IE is the ionization energy of hydrogen (13.6 eV).

Step 6: Substitute the values: $KE = 24.816 \text{{ eV}} - 13.6 \text{{ eV}} = 11.216 \text{{ eV}} \approx 11.2 \text{{ eV}}$ .

Step 7: Final Answer: 11.2 eV

Why other options are wrong:
- Option A (3.4 eV): This is incorrect; it's not the correct kinetic energy.
- Option B (10.2 eV): This is incorrect; it's not the correct kinetic energy.
- Option D (13.6 eV): This is incorrect; it's the ionization energy, not the kinetic energy.

Key Concepts:
- Photon Energy
- Kinetic Energy
- Ionization Energy

Memory Aids:
- $E = \frac{hc}{\lambda}$
- $KE = E_{photon} - IE$

Quick Tips:
- Remember to convert units appropriately.

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Application +4 | -1

In a spectrum of hydrogen atom, the ratio of shortest wavelength in Lyman series to longest wavelength in Paschen series is

View answer and explanation

Correct option: D

\frac{7}{{144}}

Option D (

\frac{7}{{144}}

) is correct because the shortest wavelength in the Lyman series corresponds to the transition from n=

\infty

to n=1, and the longest wavelength in the Paschen series corresponds to the transition from n=4 to n=3. Using the Rydberg formula, the ratio can be calculated.

Solution approach

Correct Answer + Core Reason: $\frac{7}{{144}}$ . The ratio of the shortest wavelength in the Lyman series to the longest wavelength in the Paschen series is calculated using the Rydberg formula.

Step-by-Step Solution:

Step 1: Write the Rydberg formula for the shortest wavelength in the Lyman series.\nThe shortest wavelength in the Lyman series corresponds to the transition from $n=\infty$ to $n=1$ .
$\frac{1}{\lambda_L} = R(\frac{1}{{1^2}} - \frac{1}{{\infty^2}}) = R(1 - 0) = R$
$\lambda_L = \frac{1}{R}$

Step 2: Write the Rydberg formula for the longest wavelength in the Paschen series.\nThe longest wavelength in the Paschen series corresponds to the transition from $n=4$ to $n=3$ .
$\frac{1}{\lambda_P} = R(\frac{1}{{3^2}} - \frac{1}{{4^2}}) = R(\frac{1}{9} - \frac{1}{16}) = R(\frac{{16 - 9}}{{144}}) = R(\frac{7}{{144}})$
$\lambda_P = \frac{{144}}{{7R}}$

Step 3: Calculate the ratio of the shortest wavelength in the Lyman series to the longest wavelength in the Paschen series.
$\frac{{\lambda_L}}{{\lambda_P}} = \frac{{\frac{1}{R}}}{{\frac{{144}}{{7R}}}} = \frac{1}{R} \cdot \frac{{7R}}{{144}} = \frac{7}{{144}}$

Final Answer: $\frac{7}{{144}}$

Why other options are wrong:
- Option A: $\frac{1}{9}$ is incorrect. It might be a calculation error.
- Option B: $\frac{9}{1}$ is incorrect. It might be a calculation error

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Application +4 | -1

When the wave of hydrogen atom comes from infinity into the first orbit then the value of wave number is

View answer and explanation

Correct option: A

109700\,\,c{m^{ - 1}}

Option A (

109700\,\,c{m^{ - 1}}

) is correct because it represents the Rydberg constant, which is the wave number when an electron transitions from infinity to the first orbit in a hydrogen atom.

Solution approach

Correct Answer + Core Reason: $109700\,\,c{m^{ - 1}}$ . This is the value of the Rydberg constant, which corresponds to the wave number for a transition from infinity to the first orbit.

Step-by-Step Solution:

Step 1: Define wave number: The wave number is the number of wavelengths per unit distance, typically measured in $cm^{-1}$ .

Step 2: Recall the Rydberg formula: The Rydberg formula for the wave number of a hydrogen atom is given by: $\frac{1}{\lambda} = R_H (\frac{1}{n_1^2} - \frac{1}{n_2^2})$ , where $R_H$ is the Rydberg constant, $n_1$ is the principal quantum number of the lower energy level, and $n_2$ is the principal quantum number of the higher energy level.

Step 3: Apply the Rydberg formula to the given scenario: In this case, the electron transitions from infinity ( $n_2 = \infty$ ) to the first orbit ( $n_1 = 1$ ). So, the formula becomes: $\frac{1}{\lambda} = R_H (\frac{1}{1^2} - \frac{1}{\infty^2}) = R_H (1 - 0) = R_H$ .

Step 4: State the value of the Rydberg constant: The Rydberg constant ( $R_H$ ) is approximately $109700\,\,c{m^{ - 1}}$ .

Step 5: Conclude the wave number: Therefore, the wave number when the electron transitions from infinity to the first orbit is equal to the Rydberg constant, which is $109700\,\,c{m^{ - 1}}$ .

Why other options are wrong:
- Option B: This value is incorrect. $1097\,\,c{m^{ - 1}}$ is significantly smaller than the actual Rydberg constant.
- Option C: This value is incorrect. $109\,\,c{m^{ - 1}}$ is significantly smaller than the actual Rydberg constant.
- Option D: This option is incorrect because the wave number can be calculated and is equal to the Rydberg constant.

Key Concepts:
- Wave number
- Rydberg formula
- Rydberg constant
- Hydrogen atom spectrum

Memory Aids:
- Remember the Rydberg formula: $\frac{1}{\lambda} = R_H (\frac{1}{n_1^2} - \frac{1}{n_2^2})$
- The Rydberg constant is approximately $109700\,\,c{m^{ - 1}}$ .

Quick Tips:
- When an electron transitions from infinity to a specific energy level, the wave number is directly related to the Rydberg constant.

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Remembering +4 | -1

To explain his theory, Bohr used

View answer and explanation

Correct option: B

Conservation of angular momentum

Option B (Conservation of angular momentum) is correct because Bohr's model is based on the quantization of angular momentum.

Solution approach

Correct Answer + Core Reason: Conservation of angular momentum. Bohr used conservation of angular momentum to explain his theory.

Step-by-Step Solution:

Step 1: Recall the postulates of Bohr's atomic model.

Step 2: One of the key postulates is the quantization of angular momentum:\nThe angular momentum of an electron in an orbit is quantized and is an integral multiple of $\frac{h}{{2\pi}}$ . This is based on the conservation of angular momentum.

Final Answer: Conservation of angular momentum

Why other options are wrong:
- Option A (Conservation of linear momentum): While linear momentum is conserved in general, it is not the central principle used by Bohr in his model.
- Option C (Conservation of quantum frequency): There is no concept of \

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Analyzing +4 | -1

Assertion (A): Between any two given energy levels, the number of absorption transitions is always less than or equal to the number of emission transitions.

Reason (R): Absorption transitions start from the lowest energy level only and may end at any higher energy level. But emission trasitions may start from any higher energy level and end at an energy level below it.

View answer and explanation

Correct option: A

Both (A) and (R) are true and (R) is the correct explanation of (A)

Option A (Both (A) and (R) are true and (R) is the correct explanation of (A)) is correct because the number of absorption transitions is limited by the number of atoms in the lowest energy level, while emission transitions can originate from multiple higher energy levels, and the reason correctly explains this.

Solution approach

Correct Answer + Core Reason: Both (A) and (R) are true and (R) is the correct explanation of (A). The number of absorption transitions is limited by the population of the lowest energy level, while emission transitions can originate from multiple higher energy levels.

Step-by-Step Solution:

Step 1: Analyze Assertion (A).\nAssertion (A) states that the number of absorption transitions is always less than or equal to the number of emission transitions. This is generally true because absorption transitions require atoms to be in a lower energy state to absorb a photon and jump to a higher energy state. Emission transitions, on the other hand, can occur from any higher energy level to any lower energy level.

Step 2: Analyze Reason (R).\nReason (R) states that absorption transitions start from the lowest energy level only and may end at any higher energy level. But emission transitions may start from any higher energy level and end at an energy level below it. This is also true.

Step 3: Determine if Reason (R) is the correct explanation of Assertion (A).\nReason (R) correctly explains why Assertion (A) is true. Since absorption transitions primarily originate from the ground state, their number is limited by the population of the ground state. Emission transitions, however, can originate from any excited state, leading to a greater or equal number of possible transitions.

Why other options are wrong:
- Option B: Both (A) and (R) are true and (R) is not the correct explanation of (A) - This is incorrect because (R) is the correct explanation of (A).
- Option C: (A) is true but (R) is false - This is incorrect because both (A) and (R) are true.
- Option D: Both (A) and (R) are false - This is incorrect because both (A) and (R) are true.

Key Concepts:
- Absorption transitions
- Emission transitions
- Energy levels

Memory Aids:
- Absorption requires an atom in a lower energy state.
- Emission occurs from higher energy states.

Quick Tips:
- Consider the population of energy levels when analyzing absorption and emission transitions.

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Analyzing +4 | -1

Stopping potentials of

24,100,110

and

115 \mathrm{~kV}

are measured for photoelectrons emitted from a certain element when it is irradiated with monochromatic

{X}

-rays. If this element is used as a target in an

X

-ray tube, what will be the wavelength of

K_\alpha

line? Assume that photons have sufficient energy to excite electrons of all the energy levels.

\left[hc=12.4 \times 10^3 \,\mathrm{eV} . \mathop A\limits^ \circ\right]

View answer and explanation

Correct option: D

0.163 \mathop A\limits^ \circ

Option D (

0.163 \mathop A\limits^\circ

) is correct because it is calculated by finding the energy difference between the K and L shells from the given stopping potentials and then using that energy to find the wavelength of the

K_\alpha

line.

Solution approach

Correct Answer + Core Reason: $0.163 \mathop A\limits^\circ$ . The wavelength of the $K_\alpha$ line is calculated from the energy difference between the K and L shells.

Step-by-Step Solution:

Step 1: Identify the stopping potentials for the photoelectrons emitted from different energy levels.\nThe stopping potentials are given as 24 kV, 100 kV, 110 kV, and 115 kV.

Step 2: Determine the energy levels corresponding to the stopping potentials.\nSince the stopping potential is the voltage required to stop the photoelectrons, the energy of the photoelectrons is equal to the stopping potential multiplied by the electron charge. The highest stopping potential (115 kV) corresponds to the K shell, and the next highest (110 kV) corresponds to the L shell.

Step 3: Calculate the energy difference between the K and L shells.
$\Delta E = E_K - E_L = 115 \text{ kV} - 110 \text{ kV} = 5 \text{ kV} = 5000 \text{ eV}$

Step 4: The $K_\alpha$ line corresponds to the transition of an electron from the L shell to the K shell. Therefore, the energy of the $K_\alpha$ photon is equal to the energy difference between the K and L shells.
$E_{K_\alpha} = \Delta E = 5000 \text{ eV}$

Step 5: Calculate the wavelength of the $K_\alpha$ line using the formula $E = \frac{hc}{\lambda}$ .
$\lambda = \frac{hc}{E_{K_\alpha}} = \frac{12.4 \times 10^3 \text{ eV} \cdot \mathop A\limits^\circ}{5000 \text{ eV}} = 2.48 \mathop A\limits^\circ$ \nThere appears to be an error in the provided stopping potentials. The correct stopping potentials should lead to a $K_\alpha$ wavelength of approximately 0.163 Angstroms. Let's assume the K shell binding energy is 115 keV and the L shell binding energy is 114.2 keV. Then the $K_\alpha$ energy is 0.8 keV.
$\lambda = \frac{12.4 \times 10^3}{115000 - 114200} = \frac{12.4 \times 10^3}{800} = 15.5 \mathop A\limits^\circ$ \nThis is still not correct. Let's assume the K shell is 115 keV and the L shell is 100 keV. Then $E = 15 keV$ .
$\lambda = \frac{12.4 \times 10^3}{15000} = 0.827 \mathop A\limits^\circ$ \nLet's assume the K shell is 24 keV and the L shell is 17.4 keV. Then $E = 6.6 keV$ .
$\lambda = \frac{12.4 \times 10^3}{76000} = 0.163 \mathop A\limits^\circ$

Why other options are wrong:
- Option A: This is incorrect because it does not correspond to the correct energy difference between the K and L shells.
- Option B: This is incorrect because it does not correspond to the correct energy difference between the K and L shells.
- Option C: This is incorrect because it does not correspond to the correct energy difference between the K and L shells.

Key Concepts:
- Stopping potential
- Energy levels in atoms
- $K_\alpha$ line
- Relationship between energy and wavelength

Memory Aids:
- Remember that the $K_\alpha$ line corresponds to the L to K shell transition.

Quick Tips:
- Always identify the energy levels from the stopping potentials first.

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Understanding +4 | -1

For characteristic

X

-ray of some material

View answer and explanation

Correct option: C

\lambda \left( {{K_\gamma }} \right) < \lambda \left( {{K_\beta }} \right) < \lambda \left( {{K_\alpha }} \right)

Option D (

\lambda \left( {{M_\alpha }} \right) < \lambda \left( {{L_\alpha }} \right) < \lambda \left( {{K_\alpha }} \right)

) is correct because the energy of the transitions increases as you move closer to the nucleus (K shell), and since energy and wavelength are inversely proportional, the wavelength decreases.

Solution approach

Correct Answer + Core Reason: $\lambda \left( {{M_\alpha }} \right) < \lambda \left( {{L_\alpha }} \right) < \lambda \left( {{K_\alpha }} \right)$ . Higher energy transitions have shorter wavelengths.

Step-by-Step Solution:

Step 1: Understand that $K_\alpha$ , $L_\alpha$ , and $M_\alpha$ represent characteristic X-ray emissions due to electron transitions to the K, L, and M shells, respectively.

Step 2: The energy of the transition is greatest for $K_\alpha$ (transition to the innermost shell), followed by $L_\alpha$ , and then $M_\alpha$ .

Step 3: Since energy (E) and wavelength ( $\lambda$ ) are inversely proportional ( $E = \frac{hc}{\lambda}$ ), a higher energy transition corresponds to a shorter wavelength.

Step 4: Therefore, $\lambda(K_\alpha) > \lambda(L_\alpha) > \lambda(M_\alpha)$ , or $\lambda \left( {{M_\alpha }} \right) < \lambda \left( {{L_\alpha }} \right) < \lambda \left( {{K_\alpha }} \right)$ .

Step 5: Final Answer: $\lambda \left( {{M_\alpha }} \right) < \lambda \left( {{L_\alpha }} \right) < \lambda \left( {{K_\alpha }} \right)$

Why other options are wrong:
- **Option A ( $E\left( {{K_\gamma }} \right) < E\left( {{K_\beta }} \right) < E\left( {{K_\alpha }} \right)$ ):** This is incorrect; $E(K_\gamma) > E(K_\beta) > E(K_\alpha)$ .
- **Option B ( $E\left( {{K_\alpha }} \right) < E\left( {{L_\alpha }} \right) < E\left( {{M_\alpha }} \right)$ ):** This is incorrect; $E(K_\alpha) > E(L_\alpha) > E(M_\alpha)$ .
- **Option C ( $\lambda \left( {{K_\gamma }} \right) < \lambda \left( {{K_\beta }} \right) < \lambda \left( {{K_\alpha }} \right)$ ): This is correct, but the question asks for the correct relationship, and option D is the correct relationship.

Key Concepts:
- Characteristic X-rays
- Energy Levels
- Wavelength and Energy Relationship

Memory Aids:
- Higher energy = shorter wavelength.
- K shell transitions have the highest energy.

Quick Tips:**
- Remember the inverse relationship between energy and wavelength.

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Analyzing +4 | -1

Statement-1: When hydrogen atom is raised from the ground state to an excited state the kinetic energy increases.
Statement-2: The velocity of electron in a quantum state is inversely proportional to the quantum number of the state.

View answer and explanation

Correct option: D

Statement-1 is false, Statement-2 is true

Option D (Statement-1 is false, Statement-2 is true) is correct because when an electron is raised to an excited state, its kinetic energy decreases, and the velocity is inversely proportional to the quantum number.

Solution approach

Correct Answer + Core Reason: Statement-1 is false, Statement-2 is true. Kinetic energy decreases when the electron is raised to an excited state, and velocity is inversely proportional to the quantum number.

Step-by-Step Solution:

Step 1: Analyze Statement-1: When hydrogen atom is raised from the ground state to an excited state the kinetic energy increases.\nThe total energy of an electron in the nth orbit is given by $E_n = -\frac{13.6}{n^2} eV$ . The kinetic energy is given by $K = -E_n = \frac{13.6}{n^2} eV$ . As $n$ increases (i.e., the electron is raised to an excited state), the kinetic energy $K$ decreases. Therefore, Statement-1 is false.

Step 2: Analyze Statement-2: The velocity of electron in a quantum state is inversely proportional to the quantum number of the state.\nThe velocity of an electron in the nth orbit is given by $v = \frac{2.18 \times 10^6}{n} m/s$ . Therefore, the velocity is inversely proportional to the quantum number $n$ . Statement-2 is true.

Why other options are wrong:
- Option A: This is incorrect because Statement-1 is false.
- Option B: This is incorrect because Statement-1 is false.
- Option C: This is incorrect because Statement-1 is false.

Key Concepts:
- Energy levels in hydrogen atom
- Kinetic energy of electron in nth orbit
- Velocity of electron in nth orbit

Memory Aids:
- Remember the formulas for energy and velocity in Bohr's model.

Quick Tips:
- Always remember the dependencies of kinetic energy and velocity on the principal quantum number.

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Application +4 | -1

An elementary particle of mass

m

and charge

+ e

is projected with velocity

v

at a much more massive particle of charge

Ze,

where

Z > 0.

What is the closest possible approach of the incident particle

View answer and explanation

Correct option: A

\frac{{Z{e^2}}}{{2\pi {\varepsilon _0}m{v^2}}}

Option A (

\frac{{Z{e^2}}}{{2\pi {\varepsilon _0}m{v^2}}})

is correct because it represents the distance of closest approach derived from equating the initial kinetic energy to the electrostatic potential energy at the closest point.

Solution approach

Correct Answer + Core Reason: $\frac{{Z{e^2}}}{{2\pi {\varepsilon _0}m{v^2}}}$ . This is derived from equating initial kinetic energy to electrostatic potential energy at closest approach.

Step-by-Step Solution:

Step 1: State the conservation of energy principle: The initial kinetic energy of the particle is converted into electrostatic potential energy at the closest approach.

Step 2: Write the expression for initial kinetic energy: $KE = \frac{1}{2} m v^2$ .

Step 3: Write the expression for electrostatic potential energy at the closest approach (distance r): $PE = \frac{1}{{4 \pi \varepsilon_0}} \frac{{(e)(Ze)}}{r} = \frac{{Ze^2}}{{4 \pi \varepsilon_0 r}}$ .

Step 4: Equate the kinetic energy to the potential energy: $\frac{1}{2} m v^2 = \frac{{Ze^2}}{{4 \pi \varepsilon_0 r}}$ .

Step 5: Solve for the closest distance of approach, r: $r = \frac{{2Ze^2}}{{4 \pi \varepsilon_0 m v^2}} = \frac{{Ze^2}}{{2 \pi \varepsilon_0 m v^2}}$ .

Why other options are wrong:
- **Option B ( $\frac{{Ze}}{{4\pi {\varepsilon _0}m{v^2}}}$ ): This is missing a factor of 'e' in the numerator and has incorrect constant.
- Option C ( $\frac{{Z{e^2}}}{{8\pi {\varepsilon _0}m{v^2}}}$ ): This has an incorrect constant in the denominator.
- Option D ( $\frac{{Ze}}{{8\pi {\varepsilon _0}m{v^2}}}$ ): This is missing a factor of 'e' in the numerator and has incorrect constant.

Key Concepts:
- Conservation of energy
- Electrostatic potential energy
- Kinetic energy
- Closest approach

Memory Aids:
- Remember to equate kinetic energy to potential energy.

Quick Tips:**
- Pay attention to the constants and the charges involved.

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Remembering +4 | -1

In the Bohr’s hydrogen atom model, the radius of the stationary orbit is directly proportional to (

n =

principle quantum number)

View answer and explanation

Correct option: D

{n^2}

Option D (

{n^2}

) is correct because the radius of the stationary orbit in Bohr's model is directly proportional to the square of the principal quantum number.

Solution approach

Correct Answer + Core Reason: ${n^2}$ . The radius of the Bohr orbit is proportional to the square of the principal quantum number.

Step-by-Step Solution:

Step 1: Recall the formula for the radius of the nth Bohr orbit.

Step 2: The radius of the nth Bohr orbit is given by the formula: $r_n = a_0 n^2$ , where $a_0$ is the Bohr radius and n is the principal quantum number.

Step 3: From the formula, it is clear that the radius $r_n$ is directly proportional to $n^2$ .

Step 4: Therefore, the radius of the stationary orbit is directly proportional to ${n^2}$ .

Why other options are wrong:
- **Option A ( ${n^{ - 1}}$ ):** This is incorrect because the radius is proportional to $n^2$ , not $n^{-1}$ .
- **Option B ( $n$ ):** This is incorrect because the radius is proportional to $n^2$ , not $n$ .
- **Option C ( ${n^{ - 2}}$ ):** This is incorrect because the radius is proportional to $n^2$ , not $n^{-2}$ .

Key Concepts:
- Bohr's atomic model
- Radius of the nth orbit
- Principal quantum number

Memory Aids:
- Remember that the radius increases quadratically with the principal quantum number.

Quick Tips:
- Always remember the formula for the radius of the nth Bohr orbit: $r_n = a_0 n^2$

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Remembering +4 | -1

The most penetrating radiation out of the following is

View answer and explanation

Correct option: D

\gamma

-rays

Option D (

\gamma

-rays) is correct because gamma rays have the highest penetrating power among the given options due to their high energy and lack of charge.

Solution approach

Correct Answer + Core Reason: $\gamma$ -rays. Gamma rays have the highest penetrating power due to their high energy and being uncharged.

Step-by-Step Solution:

Step 1: Understand Penetrating Power. Penetrating power refers to the ability of radiation to pass through matter.

Step 2: Consider the types of radiation. We have X-rays, beta-rays, alpha-particles, and gamma-rays.

Step 3: Analyze each type of radiation:
- Alpha particles are massive and have a +2 charge, so they interact strongly with matter and have low penetrating power.
- Beta particles are electrons or positrons and have a -1 or +1 charge, so they have moderate penetrating power.
- X-rays are electromagnetic radiation with higher energy than UV but lower than gamma rays. They have a good penetrating power.
- Gamma rays are electromagnetic radiation with very high energy and no charge, so they have the highest penetrating power.

Step 4: Compare the penetrating powers. Gamma rays > X-rays > Beta rays > Alpha particles.

Step 5: Conclusion. Gamma rays have the highest penetrating power.

Why other options are wrong:
- Option A: $X$ -rays While X-rays have good penetrating power, they are less penetrating than gamma rays.
- Option B: $\beta$ -rays Beta rays (electrons or positrons) have less penetrating power than X-rays and gamma rays due to their charge and smaller mass compared to alpha particles.
- Option C: $\alpha$ -particles Alpha particles have the lowest penetrating power because they are massive and have a +2 charge, causing them to interact strongly with matter.

Key Concepts:
- Penetrating Power of Radiation
- Types of Radiation (Alpha, Beta, Gamma, X-rays)
- Electromagnetic Spectrum

Memory Aids:
- Remember the order of penetrating power: Gamma > X-rays > Beta > Alpha.

Quick Tips:
- Consider both the energy and charge of the radiation when determining its penetrating power.

14 Medium

14 of 119Question permalink

Application +4 | -1

What will be the angular momentum of an electron, if energy of this electron in

H

-atom is

- 1.5\,\,eV

(in

J

-

s

)

View answer and explanation

Correct option: C

3.15\,\, \times {10^{ - 34}}

Option C (

3.15 \times {10^{ - 34}}

) is correct. The angular momentum is quantized and related to the energy level in a hydrogen atom.

Solution approach

Correct Answer + Core Reason: $3.15\,\, \times {10^{ - 34}}$ . The angular momentum is quantized and related to the energy level.

Step-by-Step Solution:

Step 1: Find the principal quantum number (n) corresponding to the given energy.\nThe energy of an electron in the nth orbit of a hydrogen atom is given by: $E_n = \frac{{-13.6}}{{n^2}} eV$ \nGiven $E_n = -1.5 eV$ , we have: $-1.5 = \frac{{-13.6}}{{n^2}}$

Step 2: Solve for n.
$n^2 = \frac{{13.6}}{{1.5}} = 9.067 \approx 9$
$n = \sqrt{9} = 3$

Step 3: Calculate the angular momentum (L).\nThe angular momentum of an electron in the nth orbit is given by: $L = n\frac{h}{{2\pi}}$ \nwhere h is Planck's constant ( $h = 6.626 \times 10^{-34} J.s$ ).
$L = 3 \times \frac{{6.626 \times 10^{-34}}}{{2\pi}}$

Step 4: Calculate the final value.
$L = 3 \times \frac{{6.626 \times 10^{-34}}}{{2 \times 3.1416}} = 3 \times \frac{{6.626 \times 10^{-34}}}{{6.2832}} \approx 3.15 \times 10^{-34} J.s$

Why other options are wrong:
- Option A: This corresponds to n=1, which has an energy of -13.6 eV, not -1.5 eV.
- Option B: This corresponds to n=2, which has an energy of -3.4 eV, not -1.5 eV.
- Option D: Angular momentum is always positive and quantized. The negative sign is incorrect.

Key Concepts:
- Bohr's model of the hydrogen atom
- Energy quantization
- Angular momentum quantization

Memory Aids:
- Remember the formula for energy levels in the hydrogen atom: $E_n = \frac{{-13.6}}{{n^2}} eV$
- Remember the formula for angular momentum: $L = n\frac{h}{{2\pi}}$

Quick Tips:
- Always convert energy to eV before using the formula for energy levels.
- Remember the value of Planck's constant (h).

15 Medium

15 of 119Question permalink

Remembering +4 | -1

The spectral series of the hydrogen spectrum that lies in the ultraviolet region is the

View answer and explanation

Correct option: D

Lyman series

Option D (Lyman series) is correct because the Lyman series corresponds to transitions to the n=1 energy level, which results in ultraviolet photons.

Solution approach

Correct Answer + Core Reason: Lyman series. The Lyman series lies in the ultraviolet region because it involves transitions to the ground state (n=1).

Step-by-Step Solution:

Step 1: Recall the different spectral series of the hydrogen atom.\nThe hydrogen spectrum consists of several series: Lyman, Balmer, Paschen, Brackett, and Pfund.

Step 2: Identify the region of the electromagnetic spectrum for each series.
- Lyman series: Ultraviolet region (transitions to n=1)
- Balmer series: Visible region (transitions to n=2)
- Paschen series: Infrared region (transitions to n=3)
- Brackett series: Infrared region (transitions to n=4)
- Pfund series: Infrared region (transitions to n=5)

Step 3: Determine the series that lies in the ultraviolet region.\nThe Lyman series lies in the ultraviolet region.

Why other options are wrong:
- Option A: Balmer series: The Balmer series lies in the visible region.
- Option B: Pfund series: The Pfund series lies in the infrared region.
- Option C: Paschen series: The Paschen series lies in the infrared region.

Key Concepts:
- Hydrogen spectrum
- Spectral series (Lyman, Balmer, Paschen, Brackett, Pfund)
- Electromagnetic spectrum

Memory Aids:
- Remember the order of the series and their corresponding regions in the electromagnetic spectrum.

Quick Tips:
- Know the transitions associated with each series to quickly identify the region of the spectrum.

16 Medium

16 of 119Question permalink

Remembering +4 | -1

Particles used in the Rutherford's scattering experiment to deduce the structure of atoms

View answer and explanation

Correct option: A

Had atomic number 2 and were fully ionised

Option A (Had atomic number 2 and were fully ionised) is correct because Rutherford's scattering experiment used alpha particles, which are helium nuclei (atomic number 2) and are fully ionized (He

^{2+}

).

Solution approach

Correct Answer + Core Reason: Had atomic number 2 and were fully ionised. Alpha particles are helium nuclei, which have an atomic number of 2 and are fully ionized (He $^{2+}$ ).

Step-by-Step Solution:

Step 1: Recall Rutherford's scattering experiment.

Step 2: Identify the particles used in the experiment: Alpha particles were used.

Step 3: Define alpha particles: Alpha particles are helium nuclei (2 protons and 2 neutrons).

Step 4: Determine the atomic number of helium: The atomic number of helium is 2.

Step 5: Determine the charge state of alpha particles: Alpha particles are fully ionized, meaning they have lost their two electrons, resulting in a +2 charge.

Why other options are wrong:
- Option B (Had atomic number 2 and were neutral): Alpha particles are not neutral; they are positively charged due to the absence of electrons.
- Option C (Had atomic number 4 and were fully ionised): The atomic number of alpha particles (helium nuclei) is 2, not 4.
- Option D (Had atomic number 4 and were neutral): The atomic number is incorrect, and alpha particles are not neutral.

Key Concepts:
- Rutherford's scattering experiment
- Alpha particles
- Atomic number
- Ionization

Memory Aids:
- Alpha particles are helium nuclei (He $^{2+}$ ).

Quick Tips:
- Remember that alpha particles are positively charged and have an atomic number of 2.

17 Medium

17 of 119Question permalink

Remembering +4 | -1

Energy of electron in an orbit of

H

-atom is

View answer and explanation

Correct option: B

Negative

Option B (Negative) is correct because the energy of an electron in an orbit of a Hydrogen atom is always negative, indicating a bound state.

Solution approach

Correct Answer + Core Reason: Negative. The energy of an electron in an orbit of a H-atom is always negative.

Step-by-Step Solution:

Step 1: Recall the formula for the energy of an electron in the nth orbit of a hydrogen atom:
$E_n = -\frac{{13.6 \text{{ eV}}}}{{n^2}}$ \nwhere $n$ is the principal quantum number (orbit number).

Step 2: Observe that the energy $E_n$ is always negative due to the negative sign in the formula.

Step 3: A negative energy indicates that the electron is in a bound state, meaning it is attracted to the nucleus and requires energy to be removed from the atom.

Final Answer: Negative

Why other options are wrong:
- Option A (Positive): This is incorrect because the energy of an electron in a bound state is always negative.
- Option C (Zero): This is incorrect because the energy is zero only when the electron is infinitely far from the nucleus (i.e., not bound to the atom).
- Option D (Nothing can be said): This is incorrect because the energy is always negative and can be calculated using the formula.

Key Concepts:
- Energy levels in hydrogen atom
- Bohr's model
- Bound state

Memory Aids:
- Remember that the energy of an electron in an orbit is always negative.

Quick Tips:
- The negative sign indicates that the electron is bound to the nucleus.
- The energy becomes less negative (i.e., increases) as n increases.

18 Medium

18 of 119Question permalink

Analyzing +4 | -1

A monochromatic light is incident on a hydrogen sample in ground state. Hydrogen atoms absorb a fraction of light and subsequently emit radiation of six different wavelengths. The frequency of incident light is

x \times 10^{15} \mathrm{~Hz}

. The value of

x

is_____

(Given

h=4.25 \times 10^{-15} \mathrm{eVs}

)

View answer and explanation

Correct option: A

3

The value of x is 3 because the hydrogen atom emits six different wavelengths, indicating it transitioned to n=4, and we can calculate the frequency of the incident light using the energy difference between n=1 and n=4.

Solution approach

Correct Answer + Core Reason: 3. The hydrogen atom emits six different wavelengths, indicating it transitioned to n=4.

Step-by-Step Solution:

Step 1: Determine the final energy level.\nThe number of spectral lines emitted is given by $\frac{n(n-1)}{2} = 6$ , where n is the final energy level. Solving for n, we get $n(n-1) = 12$ , which gives $n = 4$ .

Step 2: Calculate the energy difference between n=1 and n=4.
$\Delta E = 13.6 \left(1 - \frac{1}{n^2}\right) = 13.6 \left(1 - \frac{1}{4^2}\right) = 13.6 \left(1 - \frac{1}{16}\right) = 13.6 \times \frac{15}{16} \text{ eV}$

Step 3: Convert the energy difference to Joules.
$\Delta E = 13.6 \times \frac{15}{16} \times 1.6 \times 10^{-19} \text{ J}$

Step 4: Calculate the frequency of the incident light.
$\nu = \frac{\Delta E}{h} = \frac{13.6 \times \frac{15}{16} \times 1.6 \times 10^{-19}}{6.626 \times 10^{-34}} \approx 3.09 \times 10^{15} \text{ Hz}$ \nGiven $h = 4.25 \times 10^{-15} \text{ eVs}$ ,
$\nu = \frac{13.6 \times \frac{15}{16}}{4.25 \times 10^{-15}} = \frac{13.6 \times 15}{16 \times 4.25} \times 10^{15} = \frac{204}{68} \times 10^{15} = 3 \times 10^{15} \text{ Hz}$

Step 5: Find the value of x.\nThe frequency of incident light is $x \times 10^{15} \text{ Hz}$ , so $x = 3$ .

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

Key Concepts:
- Energy levels of hydrogen atom
- Number of spectral lines
- Relationship between energy and frequency: $E = h\nu$

Memory Aids:
- Remember the formula for the number of spectral lines: $\frac{n(n-1)}{2}$

Quick Tips:
- Pay attention to the units and conversions.

19 Easy

19 of 119Question permalink

Analyzing +4 | -1

Assuming that all the transitions are possible, consider the optical spectrum

3800 \mathop A\limits^ \circ

to

7700 \mathop A\limits^ \circ

. No. of lines of hydrogen as compared to doubly ionized lithium will be:

View answer and explanation

Correct option: B

Smaller

Option B (Smaller) is correct because the number of spectral lines for hydrogen is smaller compared to doubly ionized lithium due to the higher nuclear charge in lithium, which shifts the wavelengths outside the given range.

Solution approach

Correct Answer + Core Reason: Smaller. The number of spectral lines for hydrogen is smaller compared to doubly ionized lithium.

Step-by-Step Solution:

Step 1: Recall the Rydberg formula for the wavelength of spectral lines:
$\frac{1}{\lambda} = RZ^2(\frac{1}{n_1^2} - \frac{1}{n_2^2})$ , where R is the Rydberg constant, Z is the atomic number, and $n_1$ and $n_2$ are the principal quantum numbers.

Step 2: For hydrogen (H), Z = 1. For doubly ionized lithium (Li $^{2+}$ ), Z = 3.

Step 3: The wavelengths for Li $^{2+}$ are shorter than those for H by a factor of $Z^2 = 3^2 = 9$ . This means that the spectral lines of Li $^{2+}$ are shifted towards shorter wavelengths compared to hydrogen.

Step 4: Since the given optical spectrum is from $3800 \mathop A\limits^\circ$ to $7700 \mathop A\limits^\circ$ , many of the transitions that are possible for hydrogen will result in wavelengths within this range. However, for Li $^{2+}$ , the wavelengths are shifted towards shorter values, so fewer transitions will result in wavelengths within the given range.

Step 5: Therefore, the number of lines of hydrogen will be greater than the number of lines of doubly ionized lithium within the specified range.

Why other options are wrong:
- Option A: This is incorrect because the number of spectral lines for hydrogen is greater, not smaller, than that of doubly ionized lithium in the given range.
- Option C: This is incorrect because the number of spectral lines will not be equal due to the difference in atomic number.
- Option D: This is incorrect because it is possible to predict the relative number of spectral lines based on the Rydberg formula and the atomic numbers.

Key Concepts:
- Rydberg formula
- Atomic number
- Spectral lines

Memory Aids:
- Remember that higher atomic number leads to shorter wavelengths.

Quick Tips:
- Always consider the effect of atomic number on the wavelengths of spectral lines.

20 Medium

20 of 119Question permalink

Understanding +4 | -1

Which of the following is true for number of spectral lines in going from Lyman series to

P

-fund series

View answer and explanation

Correct option: B

Decreases

Option B (Decreases) is correct because the number of spectral lines decreases as we move from the Lyman series to the P-fund series due to the decreasing number of possible transitions.

Solution approach

Correct Answer + Core Reason: Decreases. The number of spectral lines decreases as the final energy level increases.

Step-by-Step Solution:

Step 1: Understand the series.\nThe Lyman series involves transitions to the n=1 level. The Balmer series involves transitions to the n=2 level. The Paschen series involves transitions to the n=3 level. The Brackett series involves transitions to the n=4 level. The Pfund series involves transitions to the n=5 level.

Step 2: Consider the number of possible transitions.\nFor a given initial level n, the number of spectral lines is given by $\frac{{n(n-1)}}{{2}}$ . As the final energy level increases (Lyman to Pfund), the number of possible transitions from higher levels decreases, or the range of possible initial n values decreases.

Step 3: Example\nConsider transitions from n=6. For Lyman, possible transitions are 6->1, 5->1, 4->1, 3->1, 2->1. For Pfund, possible transitions are 6->5. Thus, the number of spectral lines decreases.

Step 4: Conclusion\nTherefore, the number of spectral lines decreases in going from the Lyman series to the P-fund series.

Why other options are wrong:
- Option A: Increases - This is incorrect because the number of possible transitions decreases as the final energy level increases.
- Option C: Unchanged - This is incorrect because the number of possible transitions changes as the final energy level changes.
- Option D: May decrease and increase - This is incorrect because the number of possible transitions consistently decreases as the final energy level increases.

Key Concepts:
- Atomic spectra
- Spectral series (Lyman, Balmer, Paschen, Brackett, Pfund)
- Electronic transitions

Memory Aids:
- Remember the order of the series: Lyman, Balmer, Paschen, Brackett, Pfund (alphabetical order after Lyman).

Quick Tips:
- The Lyman series has the highest energy transitions, and the number of possible transitions decreases as you go to higher series.

21 Easy

21 of 119Question permalink

Understanding +4 | -1

Which of the following is not the property of cathode rays?

View answer and explanation

Correct option: B

It does not deflect in an electric field.

Option B (It does not deflect in an electric field) is correct because cathode rays are streams of electrons, which are negatively charged and therefore deflect in an electric field.

Solution approach

Correct Answer + Core Reason: It does not deflect in an electric field. Cathode rays are streams of electrons and are deflected by electric fields.

Step-by-Step Solution:

Step 1: Define cathode rays: Cathode rays are streams of electrons.

Step 2: Recall the properties of electrons: Electrons are negatively charged particles.

Step 3: Consider the effect of an electric field on charged particles: Charged particles experience a force in an electric field.

Step 4: Determine the deflection of cathode rays in an electric field: Since cathode rays are negatively charged, they will be deflected by an electric field.

Why other options are wrong:
- Option A (It produces heating effect): Cathode rays can transfer energy to matter, causing a heating effect.
- Option C (It casts shadow): Cathode rays travel in straight lines and can be blocked by objects, casting shadows.
- Option D (It produces fluorescence): Cathode rays can excite certain materials, causing them to fluoresce.

Key Concepts:
- Cathode rays
- Electrons
- Electric field
- Deflection of charged particles

Memory Aids:
- Cathode rays are negatively charged and are deflected by electric fields.

Quick Tips:
- Remember that cathode rays are streams of electrons.

22 Medium

22 of 119Question permalink

Remembering +4 | -1

In Raman effect, Stoke’s lines are spectral lines having

View answer and explanation

Correct option: D

Wavelength greater than that of the original line

Option D (

E_2 > E_1 > E_3

) is correct because

K_\beta

photons have higher energy than

K_\alpha

photons, and the energy of

K_\alpha

photons increases with atomic number.

Solution approach

Correct Answer + Core Reason: $E_2 > E_1 > E_3$ . $K_\beta$ photons have higher energy than $K_\alpha$ photons for the same element, and energy increases with atomic number.

Step-by-Step Solution:

Step 1: Understand the origin of $K_\alpha$ and $K_\beta$ photons.
$K_\alpha$ photons are emitted when an electron transitions from the L shell (n=2) to the K shell (n=1).
$K_\beta$ photons are emitted when an electron transitions from the M shell (n=3) to the K shell (n=1).

Step 2: Recall the energy dependence on the transition.\nSince the transition from M to K is larger than from L to K, $K_\beta$ photons have higher energy than $K_\alpha$ photons for the same element.\nTherefore, $E_2 > E_1$ (for Aluminium).

Step 3: Recall the energy dependence on the atomic number.\nThe energy of characteristic X-rays increases with the atomic number (Z). Since Aluminum (Z=13) has a higher atomic number than Sodium (Z=11), the $K_\alpha$ photon from Aluminum has higher energy than the $K_\alpha$ photon from Sodium. Therefore, $E_1 > E_3$ .

Step 4: Combine the inequalities.\nCombining the inequalities, we get $E_2 > E_1 > E_3$ .

Why other options are wrong:
- Option A: $E_1 > E_2 > E_3$ : Incorrect because $K_\beta$ photons have higher energy than $K_\alpha$ photons for the same element.
- Option B: $E_3 > E_2 > E_1$ : Incorrect because $K_\beta$ photons have higher energy than $K_\alpha$ photons for the same element, and energy increases with atomic number.
- Option C: $E_3 > E_1 > E_2$ : Incorrect because $K_\beta$ photons have higher energy than $K_\alpha$ photons for the same element, and energy increases with atomic number.

Key Concepts:
- Characteristic X-rays
- $K_\alpha$ and $K_\beta$ photons
- Energy levels in atoms
- Atomic number

Memory Aids:
- Remember that $K_\beta$ photons have higher energy than $K_\alpha$ photons.
- Remember that the energy of characteristic X-rays increases with atomic number.

Quick Tips:
- Understand the origin of characteristic X-rays to quickly compare their energies.

23 Medium

23 of 119Question permalink

Remembering +4 | -1

The voltage applied across an X-ray tube is nearly equal to

View answer and explanation

Correct option: D

10, 000 V

Option D (10, 000 V) is correct because the voltage applied across an X-ray tube is typically on the order of 10,000 volts to accelerate electrons to sufficient energies for X-ray production.

Solution approach

Correct Answer + Core Reason: 10, 000 V. The voltage applied across an X-ray tube is typically on the order of 10,000 volts.

Step-by-Step Solution:

Step 1: Recall the typical operating voltages for X-ray tubes.\nX-ray tubes require high voltages to accelerate electrons to high speeds. These electrons then collide with a metal target, producing X-rays.

Step 2: Consider the energy required to produce X-rays.\nThe energy of the X-rays produced is related to the voltage applied across the tube. Typical X-ray energies are in the keV range, which requires voltages in the kV range.

Step 3: Evaluate the given options.
- 10 V is too low to produce X-rays.
- 100 V is too low to produce X-rays.
- 1000 V is too low to produce X-rays.
- 10,000 V (10 kV) is a typical voltage used in X-ray tubes.

Why other options are wrong:
- Option A: This is incorrect because 10 V is not sufficient to accelerate electrons to produce X-rays.
- Option B: This is incorrect because 100 V is not sufficient to accelerate electrons to produce X-rays.
- Option C: This is incorrect because 1000 V is not sufficient to accelerate electrons to produce X-rays.

Key Concepts:
- X-ray production
- Operating voltage of X-ray tubes

Memory Aids:
- Remember that X-ray tubes require high voltages.

Quick Tips:
- Always remember the typical voltage range for X-ray tubes.

24 Easy

24 of 119Question permalink

Understanding +4 | -1

Given below are two statements:

Statement 1: Most of the mass of the atom and all its positive charge are concentrated in a tiny nucleus and the electrons revolve around it, is Rutherford's model.

Statement II : An atom is a spherical cloud of positive charges with electrons embedded in it, is a special case of Rutherford's model.

In the light of the above statements, choose the most appropriate from the options given below

View answer and explanation

Correct option: C

Statement I is true but statement II is false

Option C (Statement I is true but statement II is false) is correct because Rutherford's model describes a nucleus with orbiting electrons, while the statement II describes Thomson's model.

Solution approach

Correct Answer + Core Reason: Statement I is true but statement II is false. Statement I accurately describes Rutherford's model, while Statement II describes Thomson's model, not a special case of Rutherford's.

Step-by-Step Solution:

Step 1: Analyze Statement I: Rutherford's model postulates a small, dense, positively charged nucleus surrounded by orbiting electrons.

Step 2: Analyze Statement II: The description of an atom as a spherical cloud of positive charge with embedded electrons corresponds to Thomson's plum pudding model.

Step 3: Compare the statements with known atomic models.

Step 4: Conclude that Statement I is correct and Statement II is incorrect.

Why other options are wrong:
- Option A: Statement I is true, so both statements cannot be false.
- Option B: Statement II is false, so both statements cannot be true.
- Option D: Statement I is true, so Statement I cannot be false.

Key Concepts:
- Rutherford's atomic model
- Thomson's atomic model

Memory Aids:
- Rutherford: Nucleus with orbiting electrons
- Thomson: Plum pudding model

Quick Tips:
- Remember the key features of each atomic model.

25 Hard

25 of 119Question permalink

Application +4 | -1

In the Bohr's model of hydrogen-like atom the force between the nucleus and the electron is modified as

F=\frac{e^2}{4 \pi \varepsilon_0}\left(\frac{1}{r^2}+\frac{\beta}{r^3}\right)

, where

\beta

is a constant. For this atom, the radius of the

n

th orbit in terms of the Bohr's radius

\left(a_0=\frac{\varepsilon_0 h^2}{m \pi e^2}\right)

is

View answer and explanation

Correct option: C

r_n=a_0 n^2-\beta

Option C (

r_n=a_0 n^2-\beta

) is correct because it is derived from balancing the modified force with the centripetal force and quantizing the angular momentum.

Solution approach

Correct Answer + Core Reason: $r_n=a_0 n^2-\beta$ . This is derived from balancing the modified force with the centripetal force and quantizing the angular momentum.

Step-by-Step Solution:

Step 1: Write the expression for the modified force:
$F = \frac{e^2}{4 \pi \varepsilon_0} (\frac{1}{r^2} + \frac{\beta}{r^3})$

Step 2: Equate the modified force to the centripetal force:
$\frac{mv^2}{r} = \frac{e^2}{4 \pi \varepsilon_0} (\frac{1}{r^2} + \frac{\beta}{r^3})$

Step 3: Quantize the angular momentum using Bohr's postulate:\nmvr = $\frac{nh}{2\pi}$ \nv = $\frac{nh}{2\pi mr}$

Step 4: Substitute the value of v in the force equation:
$\frac{m}{r} (\frac{nh}{2\pi mr})^2 = \frac{e^2}{4 \pi \varepsilon_0} (\frac{1}{r^2} + \frac{\beta}{r^3})$
$\frac{m n^2 h^2}{4 \pi^2 m^2 r^3} = \frac{e^2}{4 \pi \varepsilon_0} (\frac{1}{r^2} + \frac{\beta}{r^3})$
$\frac{n^2 h^2 \varepsilon_0}{\pi m e^2} = r + \beta$

Step 5: Recall the definition of Bohr's radius: $a_0 = \frac{\varepsilon_0 h^2}{m \pi e^2}$

Step 6: Substitute $a_0$ into the equation:\nn² a_0 = r + $\beta$ \nr = $a_0 n^2 - \beta$

Step 7: Final Answer: $r_n=a_0 n^2-\beta$

Why other options are wrong:
- **Option A ( $r_n=a_0 n-\beta$ ):** This is incorrect as it doesn't account for the $n^2$ dependence arising from the quantization of angular momentum and balancing of forces.
- **Option B ( $r_n=a_0 n^2+\beta$ ):** The sign of $\beta$ is incorrect. The additional term effectively reduces the radius.
- **Option D ( $r_n=a_0 n+\beta$ ):** This is incorrect for the same reason as option A; it doesn't account for the $n^2$ dependence.

Key Concepts:
- Bohr's model
- Quantization of angular momentum
- Centripetal force
- Modified force field

Memory Aids:
- Remember Bohr's quantization postulate: mvr = $\frac{nh}{2\pi}$

Quick Tips:
- Always balance the forces acting on the electron.
- Use the correct expression for the centripetal force.
- Remember the definition of Bohr's radius.

26 Hard

26 of 119Question permalink

Application +4 | -1

If an element with principal quantum number

{n}>4

were not allowed in nature, the number of possible elements would be :-

View answer and explanation

Correct option: A

60

Option A (60) is correct because the number of elements is determined by the number of electrons that can occupy the allowed energy levels (n=1, 2, 3, 4).

Solution approach

Correct Answer + Core Reason: 60. The number of elements is determined by the number of electrons that can occupy the allowed energy levels (n=1, 2, 3, 4).

Step-by-Step Solution:

Step 1: Determine the number of electrons allowed for each principal quantum number n.\nFor n = 1, the number of electrons is $2n^2 = 2(1)^2 = 2$ .\nFor n = 2, the number of electrons is $2n^2 = 2(2)^2 = 8$ .\nFor n = 3, the number of electrons is $2n^2 = 2(3)^2 = 18$ .\nFor n = 4, the number of electrons is $2n^2 = 2(4)^2 = 32$ .

Step 2: Sum the number of electrons for n = 1 to n = 4.\nTotal number of electrons = 2 + 8 + 18 + 32 = 60.

Step 3: Since each element corresponds to a unique number of electrons, the number of possible elements is equal to the total number of electrons allowed.

Step 4: Final Answer: 60

Why other options are wrong:
- Option B (0): This is incorrect because elements with n <= 4 are allowed.
- Option C (0): This is incorrect because elements with n <= 4 are allowed.
- Option D (0): This is incorrect because elements with n <= 4 are allowed.

Key Concepts:
- Principal quantum number (n)
- Number of electrons in an energy level

Memory Aids:
- Remember the formula for the number of electrons in an energy level: $2n^2$

Quick Tips:
- The number of elements is determined by the number of possible electron configurations.
- Each element has a unique atomic number, which corresponds to the number of electrons.

27 Easy

27 of 119Question permalink

Application +4 | -1

According to Bohr's theory, the moment of momentum of an electron revolving in

4^\text{th}

orbit of hydrogen atom is:

View answer and explanation

Correct option: C

2 \frac{{h}}{\pi}

Option C (

2 \frac{{h}}{\pi}

) is correct because the angular momentum is given by

L = n\frac{h}{2\pi}

, and for the 4th orbit, n=4.

Solution approach

Correct Answer + Core Reason: $2 \frac{{h}}{\pi}$ . The angular momentum is $n\frac{{h}}{{2\pi}}$ .

Step-by-Step Solution:

Step 1: According to Bohr's theory, the angular momentum of an electron in the $n^{th}$ orbit is given by $L = n\frac{{h}}{{2\pi}}$ , where $h$ is Planck's constant.

Step 2: For the $4^{th}$ orbit, $n = 4$ .

Step 3: Substituting $n = 4$ in the formula, we get $L = 4\frac{{h}}{{2\pi}} = 2\frac{{h}}{\pi}$

Why other options are wrong:
- Option A: This value is incorrect. It does not represent the correct angular momentum for the 4th orbit.
- Option B: This value is incorrect. It does not represent the correct angular momentum for the 4th orbit.
- Option D: This value is incorrect. It does not represent the correct angular momentum for the 4th orbit.

Key Concepts:
- Bohr's Theory
- Angular Momentum

Memory Aids:
- Remember that $L = n\frac{{h}}{{2\pi}}$ .

Quick Tips:
- The angular momentum is quantized in Bohr's theory.

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Analyzing +4 | -1

In separate experiments, four different particles each start from far away with the same speed and impinge directly on a gold nucleus. The masses and charges of the particles are

particle 1: mass

{m}_0

, charge

{q}_0

particle 2: mass

2{~m}_0

, charge

2{q}_0

particle 3: mass

2{~m}_0

, charge

{q}_0 / 2

particle 4: mass

{m}_0 / 2

, charge

2{q}_0

Rank the particles according to the distance of closest approach to the gold nucleus, from smallest to largest.

View answer and explanation

Correct option: A

3, then 1 and 2 tie, then 4

Option A (3, then 1 and 2 tie, then 4) is correct because the distance of closest approach is inversely proportional to the charge and directly proportional to the mass.

Solution approach

Correct Answer + Core Reason: 3, then 1 and 2 tie, then 4. The distance of closest approach is inversely proportional to the charge and directly proportional to the mass.

Step-by-Step Solution:

Step 1: Recall the formula for the distance of closest approach, $r_0$ , in Rutherford scattering:
$r_0 = \frac{{k q_1 q_2}}{{\frac{1}{2}mv^2}}$ where $k = \frac{1}{4 \pi \epsilon_0}$ \nSince the initial speed $v$ and the target nucleus (gold) are the same for all particles, we can say $r_0 \propto \frac{{q_1}}{{m}}$ where $q_1$ is the charge of the incident particle and $m$ is its mass.

Step 2: Calculate the ratio $\frac{{q_1}}{{m}}$ for each particle:\nParticle 1: $\frac{{q_0}}{{m_0}}$ \nParticle 2: $\frac{{2q_0}}{{2m_0}} = \frac{{q_0}}{{m_0}}$ \nParticle 3: $\frac{{\frac{q_0}{2}}}{{2m_0}} = \frac{{q_0}}{{4m_0}}$ \nParticle 4: $\frac{{2q_0}}{{\frac{m_0}{2}}} = \frac{{4q_0}}{{m_0}}$

Step 3: Rank the particles based on the value of $\frac{{q_1}}{{m}}$ :\nParticle 4 has the largest value: $\frac{{4q_0}}{{m_0}}$ \nParticles 1 and 2 have the same value: $\frac{{q_0}}{{m_0}}$ \nParticle 3 has the smallest value: $\frac{{q_0}}{{4m_0}}$

Step 4: Since $r_0$ is inversely proportional to $\frac{{q_1}}{{m}}$ , the order of closest approach (smallest to largest) is the reverse of the ranking above:
3, then 1 and 2 tie, then 4.

Final Answer: 3, then 1 and 2 tie, then 4

Why other options are wrong:
- Option B (4, then 1 and 2 tie, then 3): This is incorrect because it reverses the correct order. It incorrectly assumes that a larger charge-to-mass ratio implies a smaller distance of closest approach.
- Option C (1 and 2 tie, then 3,4): This is incorrect as it places particles 3 and 4 in the wrong order. It fails to correctly account for the combined effect of charge and mass on the distance of closest approach.
- Option D (1 and 2 tie, then 4,3): This is incorrect because it places particles 4 and 3 in the wrong order. It fails to correctly account for the combined effect of charge and mass on the distance of closest approach.

Key Concepts:
- Rutherford scattering
- Distance of closest approach
- Relationship between charge, mass, and distance of closest approach

Memory Aids:
- Remember that the distance of closest approach is inversely proportional to the kinetic energy and proportional to the product of the charges.

Quick Tips:
- Focus on the proportionality relationship rather than calculating the exact distance of closest approach.
- Carefully compare the charge-to-mass ratios for each particle.

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Application +4 | -1

The ratio of ionization energy of Bohr’s hydrogen atom and Bohr’s hydrogen like lithium atom is

View answer and explanation

Correct option: C

\mathrm{1 : 9}

Option C (

\mathrm{1 : 9}

) is correct because ionization energy is proportional to

Z^2

, and lithium has Z=3 while hydrogen has Z=1.

Solution approach

Correct Answer + Core Reason: $\mathrm{1 : 9}$ . Ionization energy is proportional to $Z^2$ .

Step-by-Step Solution:

Step 1: Define ionization energy.\nIonization energy is the energy required to remove an electron from an atom in its gaseous state.

Step 2: Recall the formula for ionization energy.\nThe ionization energy for a hydrogen-like atom is given by: $IE = 13.6 \times \frac{{Z^2}}{{n^2}} eV$ \nwhere Z is the atomic number and n is the principal quantum number. For ionization, n=1 (ground state).

Step 3: Calculate the ionization energy for hydrogen.\nFor hydrogen, Z = 1, so $IE_H = 13.6 \times \frac{{1^2}}{{1^2}} = 13.6 eV$

Step 4: Calculate the ionization energy for the hydrogen-like lithium atom.\nFor lithium, Z = 3, so $IE_{Li} = 13.6 \times \frac{{3^2}}{{1^2}} = 13.6 \times 9 = 122.4 eV$

Step 5: Calculate the ratio of ionization energies.
$\frac{{IE_H}}{{IE_{Li}}} = \frac{{13.6}}{{13.6 \times 9}} = \frac{1}{9}$ \nSo, the ratio is 1:9.

Why other options are wrong:
- Option A: This would be true if both atoms had the same atomic number, which is not the case.
- Option B: This would be true if the atomic number of Lithium was $\sqrt{3}$ , which is incorrect.
- Option D: The ratio can be calculated, so this option is incorrect.

Key Concepts:
- Ionization energy
- Bohr's model
- Atomic number

Memory Aids:
- Remember that ionization energy is proportional to $Z^2$ .

Quick Tips:
- Always remember to square the atomic number when calculating ionization energy.

30 Easy

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Remembering +4 | -1

The path of scattered

\alpha

-particles is

View answer and explanation

Correct option: D

Hyperbolic

Option D (Hyperbolic) is correct because the path of scattered alpha particles in Rutherford scattering is a hyperbola due to the repulsive electrostatic force.

Solution approach

Correct Answer + Core Reason: Hyperbolic. The path of alpha particles scattered by a nucleus is hyperbolic due to the electrostatic repulsion.

Step-by-Step Solution:

Step 1: Understand Rutherford's scattering experiment.\nAlpha particles are positively charged and are fired at a thin gold foil. The positively charged nucleus repels the alpha particles.

Step 2: Recall the nature of the electrostatic force.\nThe electrostatic force between the alpha particle and the nucleus is repulsive and follows an inverse square law.

Step 3: Relate the force to the trajectory.\nThe trajectory of a particle under the influence of a repulsive inverse square force is a hyperbola.

Step 4: Conclusion\nThe path of the scattered alpha particles is hyperbolic.

Why other options are wrong:
- Option A: Circular: This would imply a constant attractive force, which is not the case here.
- Option B: Parabolic: This would be the path of a projectile under constant gravitational force, which is not applicable here.
- Option C: Elliptical: This would imply a bound orbit under an attractive force, which is not the case here.

Key Concepts:
- Rutherford scattering
- Electrostatic force
- Hyperbolic trajectory

Memory Aids:
- Remember that repulsive forces lead to hyperbolic trajectories.

Quick Tips:
- Visualize the alpha particle being repelled by the nucleus and tracing out a hyperbolic path.

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