The atomic transition gives rise to the radiation of frequency ${10^4}\;{\rm{MHz}}$ The change in energy per mole of atoms taking place would be

$6.62 \times {10^{ - 30}}{\rm{J}}$

$5.32 \times {10^{ - 28}}{\rm{J}}$

$6.62 \times {10^{ - 20}}{\rm{J}}$

The measurement of the electron position is associated with an uncertainty in momentum, which is equal to $1 \times {10^{ - 18}}\;{\rm{g\;cm\;}}{{\rm{s}}^{ - 1}}.$. The uncertainty in electron velocity is: (mass of an electron is$9 \times {10^{ - 28}}\;{\rm{g}}$)

$1.1 \times {10^9}{\rm{cm\;}}{{\rm{s}}^{ - 1}}$

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

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

$1 \times {10^{11}}{\rm{cm\;}}{{\rm{s}}^{ - 1}}$

STRUCTURE OF ATOM MCQs for NEET UG CHEMISTRY — Medium, Page 5

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Understanding +4 | -1

Which of the following statements does not form a part of Bohr’s model of hydrogen atom?

View answer and explanation

Correct option: D

The position and velocity of the electrons in the orbit cannot be determined simultaneously

Option D (The position and velocity of the electrons in the orbit cannot be determined simultaneously) is correct because this statement refers to the Heisenberg Uncertainty Principle, which is not part of Bohr's model. Bohr's model assumes that electrons have definite positions and velocities in their orbits.

Solution approach

Correct Answer + Core Reason: The position and velocity of the electrons in the orbit cannot be determined simultaneously. This statement is related to Heisenberg's Uncertainty Principle, which is not part of Bohr's model.

Step-by-Step Solution:

Step 1: State the postulates of Bohr's model:
- Electrons revolve around the nucleus in specific orbits.
- The energy of electrons in these orbits is quantized.
- Electrons can only transition between orbits by absorbing or emitting energy.
- The electron in the orbit nearest the nucleus has the lowest energy.

Step 2: State the Heisenberg Uncertainty Principle: The position and velocity of an electron cannot be simultaneously determined with perfect accuracy.

Step 3: Compare Bohr's model with the Uncertainty Principle: Bohr's model assumes that electrons have definite positions and velocities, which contradicts the Uncertainty Principle.

Step 4: Final Answer: The statement that does not form a part of Bohr's model is that the position and velocity of the electrons in the orbit cannot be determined simultaneously.

Why other options are wrong:
- Option A: This is a postulate of Bohr's model.
- Option B: This is a postulate of Bohr's model.
- Option C: This is a postulate of Bohr's model.

Key Concepts:
- Bohr's model
- Heisenberg Uncertainty Principle
- Quantization of energy

Memory Aids:
- Bohr's model is a classical model with defined orbits, unlike quantum mechanics.

Quick Tips:
- Remember the key postulates of Bohr's model.

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Analyzing +4 | -1

Which of the following electron transition in hydrogen atom will require largest amount of energy?

View answer and explanation

Correct option: A

From n=1 to n=2

Option A (From n=1 to n=2) is correct because the largest amount of energy is required for transitions starting from the ground state (n=1).

Solution approach

Correct Answer + Core Reason: From n=1 to n=2. The largest energy is required for transitions originating from the ground state.

Step-by-Step Solution:

Step 1: Recall the energy level formula for the hydrogen atom: $E_n = -\frac{13.6 eV}{n^2}$ .

Step 2: The energy required for a transition is given by the difference in energy levels: $\Delta E = E_{final} - E_{initial} = -13.6 eV (\frac{1}{n_{final}^2} - \frac{1}{n_{initial}^2})$ .

Step 3: Calculate the energy required for each transition:
- A: From n=1 to n=2: $\Delta E = -13.6 eV (\frac{1}{2^2} - \frac{1}{1^2}) = -13.6 eV (\frac{1}{4} - 1) = -13.6 eV (-\frac{3}{4}) = 10.2 eV$
- B: From n=2 to n=3: $\Delta E = -13.6 eV (\frac{1}{3^2} - \frac{1}{2^2}) = -13.6 eV (\frac{1}{9} - \frac{1}{4}) = -13.6 eV (-\frac{5}{36}) = 1.89 eV$
- C: From $n = \infty$ to n=1: $\Delta E = -13.6 eV (\frac{1}{1^2} - \frac{1}{\infty^2}) = -13.6 eV (1 - 0) = -13.6 eV$ . This is an emission, and the question asks for the largest amount of energy required (absorption).
- D: From n=3 to n=5: $\Delta E = -13.6 eV (\frac{1}{5^2} - \frac{1}{3^2}) = -13.6 eV (\frac{1}{25} - \frac{1}{9}) = -13.6 eV (-\frac{16}{225}) = 0.97 eV$

Step 4: Compare the energy values. The largest energy required is 10.2 eV for the transition from n=1 to n=2.

Why other options are wrong:
- Option B: From n=2 to n=3 requires less energy than from n=1 to n=2.
- Option C: From $n = \infty$ to n=1 is an emission, not an absorption, and it releases the most energy, but the question asks for the largest amount of energy *required* (absorption).
- Option D: From n=3 to n=5 requires less energy than from n=1 to n=2.

Key Concepts:
- Energy Levels in Hydrogen Atom
- Rydberg Formula
- Energy Transitions

Memory Aids:
- The energy difference between levels decreases as n increases.
- Transitions to or from n=1 involve the largest energy changes.

Quick Tips:
- Remember that transitions to higher energy levels (absorption) require energy, while transitions to lower energy levels (emission) release energy.
- Transitions involving n=1 generally require the most energy.

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What does the electronic configuration

1{s^2},2{s^2},2{p^5},3{s^1}

indicate?

View answer and explanation

Correct option: C

Excited state of neon

Option C (Excited state of neon) is correct because neon's ground state configuration is 1s²2s²2p⁶, and the given configuration shows an electron promoted from 2p to 3s.

Solution approach

Correct Answer + Core Reason: Excited state of neon. The electronic configuration indicates an excited state of neon because an electron has been promoted to a higher energy level.

Step-by-Step Solution:

Step 1: Write the ground state electronic configurations.\nFluorine (F): 1s² 2s² 2p⁵\nNeon (Ne): 1s² 2s² 2p⁶\nO₂⁻: (17 electrons) 1s² 2s² 2p⁶ 3s² 3p⁵

Step 2: Compare the given configuration with the ground state configurations.\nThe given configuration is 1s² 2s² 2p⁵ 3s¹.\nThis configuration has one electron moved from the 2p orbital to the 3s orbital.

Step 3: Determine which species corresponds to the excited state.\nThis configuration has the same number of electrons as neon (10 electrons), but it is in an excited state because the 3s orbital is occupied.

Why other options are wrong:
- Option A: The ground state of fluorine is 1s² 2s² 2p⁵, which is different from the given configuration.
- Option B: The given configuration is not the ground state of fluorine, but an excited state of neon.
- Option D: The number of electrons in O₂⁻ is 17, which is different from the given configuration (10 electrons).

Key Concepts:
- Electronic configuration
- Ground state
- Excited state

Memory Aids:
- Remember that an excited state has electrons in higher energy levels than the ground state.

Quick Tips:
- Compare the given configuration with the ground state configurations of the given species.

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Understanding +4 | -1

The representation of the ground state electronic configuration of He by box-diagram as is wrong because it violates

View answer and explanation

Correct option: C

Pauli exclusion principle

Option C (Pauli exclusion principle) is correct because the box diagram showing both electrons in the same box with the same spin violates the Pauli exclusion principle.

Solution approach

Correct Answer + Core Reason: Pauli exclusion principle. The Pauli exclusion principle states that no two electrons in an atom can have the same set of four quantum numbers.

Step-by-Step Solution:

Step 1: Identify the electronic configuration of He. Helium has two electrons.

Step 2: Consider the box diagram representation. If both electrons are in the same box, they have the same spin quantum number.

Step 3: Apply the Pauli exclusion principle. This violates the Pauli exclusion principle because both electrons would have the same four quantum numbers (n, l, ml, and ms).

Step 4: Conclusion: The representation violates the Pauli exclusion principle.

Why other options are wrong:
- Option A: Heisenberg’s uncertainty principle: This principle deals with the uncertainty in determining position and momentum simultaneously, not the arrangement of electrons in orbitals.
- Option B: Bohr’s quantization theory of angular momenta: This theory relates to the quantization of angular momentum in Bohr's model, not the filling of orbitals.
- Option D: Hund’s rule: Hund's rule states that electrons will individually occupy each orbital within a subshell before doubly occupying any one orbital. This applies to multi-electron systems with degenerate orbitals, not helium.

Key Concepts:
- Pauli Exclusion Principle
- Electronic Configuration
- Quantum Numbers

Memory Aids:
- Pauli: No two electrons can have the same address (quantum numbers).

Quick Tips:
- Remember the Pauli exclusion principle when filling orbitals.

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Remembering +4 | -1

For a given value of azimuthal quantum number

l

, the total number of values for the magnetic quantum number m are given by:

View answer and explanation

Correct option: B

2l + 1

Option B is correct because for a given value of l, the magnetic quantum number m can take 2l+1 values ranging from -l to +l.

Solution approach

Correct Answer + Core Reason: $2l + 1$ . The total number of values for the magnetic quantum number m is $2l+1$ .

Step-by-Step Solution:

Step 1: Recall the relationship between the azimuthal quantum number (l) and the magnetic quantum number (m).
- For a given value of the azimuthal quantum number (l), the magnetic quantum number (m) can take integer values from -l to +l, including 0.

Step 2: Determine the number of possible values for m.
- The total number of values for m is given by $2l + 1$ .
- For example, if $l = 0$ , then $m = 0$ , and there is $2(0) + 1 = 1$ value.
- If $l = 1$ , then $m = -1, 0, +1$ , and there are $2(1) + 1 = 3$ values.
- If $l = 2$ , then $m = -2, -1, 0, +1, +2$ , and there are $2(2) + 1 = 5$ values.

Why other options are wrong:
- Option A: $l + 1$ is incorrect.
- Option C: $2l - 1$ is incorrect.
- Option D: $l + 2$ is incorrect.

Key Concepts:
- Azimuthal quantum number (l)
- Magnetic quantum number (m)

Memory Aids:
- Remember the formula $2l + 1$ for the number of possible values of m.

Quick Tips:
- The magnetic quantum number determines the spatial orientation of an orbital.

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Application +4 | -1

The atomic transition gives rise to the radiation of frequency

{10^4}\;\mathrm{MHz}

The change in energy per mole of atoms taking place would be

View answer and explanation

Correct option: D

3.99 J

Option D is correct. The change in energy per mole is calculated by multiplying the energy of a single photon by Avogadro's number, resulting in approximately 3.99 J.

Solution approach

Correct Answer + Core Reason: 3.99 J. The change in energy per mole is calculated by multiplying the energy of one photon by Avogadro's number.

Step-by-Step Solution:

Step 1: Calculate the energy of a single photon: $E = hf$ , where $h = 6.626 \times 10^{-34} \text{ Js}$ and $f = 10^4 \text{ MHz} = 10^{10} \text{ Hz}$ .

Step 2: Substitute the values: $E = (6.626 \times 10^{-34} \text{ Js})(10^{10} \text{ Hz}) = 6.626 \times 10^{-24} \text{ J}$ .

Step 3: Calculate the energy per mole: $E_{\text{mole}} = E \times N_A$ , where $N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$ .

Step 4: Substitute the values: $E_{\text{mole}} = (6.626 \times 10^{-24} \text{ J})(6.022 \times 10^{23} \text{ mol}^{-1}) = 3.99 \text{ J/mol}$ .

Why other options are wrong:
- Option A: $6.62 \times {10^{ - 30}}\mathrm{J}$ : Incorrect; this is a much smaller value and likely a calculation error.
- Option B: $5.32 \times {10^{ - 28}}\mathrm{J}$ : Incorrect; this is a much smaller value and likely a calculation error.
- Option C: $6.62 \times {10^{ - 20}}\mathrm{J}$ : Incorrect; this is the energy of a single photon multiplied by 10⁴ instead of Avogadro's number.

Key Concepts:
- Energy of a photon
- Frequency
- Avogadro's number

Memory Aids:
- Remember the formula $E = hf$ and how to calculate energy per mole.

Quick Tips:
- Ensure units are consistent before calculating. Use the correct values for Planck's constant and Avogadro's number.

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Analyzing +4 | -1

Assertion (A): Red light of any brightness may shine on a piece of potassium-metal but no photoelectrons are ejected but as soon as even a very weak yellow light shines on the potassium-metal, the photoelectric effect is observed.
Reason (R): The number of electrons ejected does depend upon the brightness of light and not on the kinetic energy of the ejected electrons.

View answer and explanation

Correct option: A

Both Assertion and Reason are correct; Reason is the correct explanation of Assertion

Option A (Both Assertion and Reason are correct; Reason is the correct explanation of Assertion) is correct because the photoelectric effect depends on the frequency of light, not brightness, and the number of ejected electrons depends on brightness.

Solution approach

Correct Answer + Core Reason: Both Assertion and Reason are correct; Reason is the correct explanation of Assertion. The assertion correctly describes the photoelectric effect, and the reason correctly explains the dependence of electron ejection on frequency and brightness.

Step-by-Step Solution:

Step 1: Analyze the Assertion: The assertion describes the photoelectric effect. Red light, regardless of brightness, may not eject photoelectrons from potassium because its frequency is below the threshold frequency. Yellow light, even if weak, can eject photoelectrons because its frequency is above the threshold frequency.

Step 2: Analyze the Reason: The reason states that the number of electrons ejected depends on the brightness (intensity) of light, while the kinetic energy of the ejected electrons depends on the frequency of light. This is a correct statement according to the laws of the photoelectric effect.

Step 3: Determine the relationship between Assertion and Reason: The reason correctly explains the assertion. The frequency of light determines whether photoelectrons are ejected, and the brightness determines the number of electrons ejected.

Why other options are wrong:
- Option B: Both Assertion and Reason are correct; Reason is not the correct explanation of Assertion. This is wrong because the reason is the correct explanation.
- Option C: Assertion is correct; Reason is incorrect. This is wrong because both assertion and reason are correct.
- Option D: Assertion is incorrect; Reason is correct. This is wrong because the assertion is correct.

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

Memory Aids:
- Photoelectric effect depends on frequency, not intensity, for electron ejection.
- Intensity determines the number of electrons ejected.

Quick Tips:
- Remember that the energy of a photon is given by $E = hf$ , where $h$ is Planck's constant and $f$ is the frequency.

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Understanding +4 | -1

Photoelectric effect is the phenomenon in which:

View answer and explanation

Correct option: D

Electrons come out of a metal with different velocities not greater than a certain value which depends only on the frequency of the incident light wave and not on its intensity

Option D (Electrons come out of a metal with different velocities not greater than a certain value which depends only on the frequency of the incident light wave and not on its intensity) is correct because the kinetic energy of emitted electrons depends on the frequency of the incident light, and there's a maximum kinetic energy.

Solution approach

Correct Answer + Core Reason: Electrons come out of a metal with different velocities not greater than a certain value which depends only on the frequency of the incident light wave and not on its intensity. This accurately describes the photoelectric effect.

Step-by-Step Solution:

Step 1: Define the photoelectric effect.
- The photoelectric effect is the emission of electrons from a metal surface when light of sufficient frequency shines on it.

Step 2: Explain the key characteristics of the photoelectric effect:
- The kinetic energy of the emitted electrons depends on the frequency of the incident light.
- There is a threshold frequency below which no electrons are emitted, regardless of the intensity of the light.
- The intensity of the light affects the number of electrons emitted (photocurrent), but not their kinetic energy.
- The emitted electrons have a range of velocities, up to a maximum value determined by the frequency of the incident light and the work function of the metal.

Why other options are wrong:
- Option A: Photons come out of a metal when it is hit by a beam of electrons is incorrect because this describes the inverse photoelectric effect or Bremsstrahlung, not the photoelectric effect.
- Option B: Photons come out of the nucleus of an atom under the action of an electric field is incorrect because this is not related to the photoelectric effect.
- Option C: Electrons come out of a metal with a constant velocity which depends on the frequency and intensity of incident light wave is incorrect because the electrons have a range of velocities, and the velocity depends on frequency but not intensity.

Key Concepts:
- Photoelectric effect
- Threshold frequency
- Work function
- Kinetic energy of emitted electrons
- Intensity of light

Memory Aids:
- Remember that frequency affects the energy of electrons, while intensity affects the number of electrons.

Quick Tips:
- The photoelectric effect provides evidence for the particle nature of light (photons).

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An iron rod when heated emits radiation with following colour change on heating:

View answer and explanation

Correct option: A

dull red to red to white and finally blue

Option A (dull red to red to white and finally blue) is correct because as temperature increases, the emitted radiation shifts towards shorter wavelengths, progressing from red to blue.

Solution approach

Correct Answer + Core Reason: dull red to red to white and finally blue. This is due to Wien's displacement law, which states that the peak wavelength of emitted radiation is inversely proportional to temperature.

Step-by-Step Solution:

Step 1: Understand the relationship between temperature and color of emitted radiation.

Step 2: As an object is heated, it starts to glow. The color changes with increasing temperature.

Step 3: Initially, at lower temperatures, the object emits primarily infrared radiation, which is not visible. As the temperature increases, it starts to emit red light (dull red).

Step 4: With further increase in temperature, the color shifts to red, then to orange, yellow, white, and finally blue at very high temperatures.

Why other options are wrong:
- Option B: Red, dull red, blue and finally white - This is incorrect because the order of color change with increasing temperature is from red towards blue, not the other way around.
- Option C: dull red, white , red , blue - This is incorrect because the color change is continuous with increasing temperature.
- Option D: dull red and then red only - This is incorrect because the color continues to change with increasing temperature beyond red.

Key Concepts:
- Blackbody radiation
- Wien's displacement law
- Relationship between temperature and color

Memory Aids:
- Remember the order of colors in the visible spectrum (ROYGBIV) and how it relates to increasing energy and temperature.

Quick Tips:
- Higher temperature means shorter wavelength (bluer color).

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Assertion: A black body is also a perfect radiator of radiant energy.

Reason: A black body is in thermal equilibrium with its surroundings.

View answer and explanation

Correct option: B

If both A and R are true but R is not the correct explanation of A

Option B (If both A and R are true but R is not the correct explanation of A) is correct because while both statements are true, thermal equilibrium doesn't directly explain why a black body is a perfect radiator.

Solution approach

Correct Answer + Core Reason: If both A and R are true but R is not the correct explanation of A. A black body is indeed a perfect radiator, and it is in thermal equilibrium with its surroundings, but the equilibrium doesn't explain the radiating property.

Step-by-Step Solution:

Step 1: Analyze the assertion: A black body is a perfect radiator. This is a fundamental property of black bodies.

Step 2: Analyze the reason: A black body is in thermal equilibrium with its surroundings. This is also true; any object in a stable environment will eventually reach thermal equilibrium.

Step 3: Determine if the reason explains the assertion. Thermal equilibrium means the body is at the same temperature as its surroundings and is exchanging energy at the same rate it absorbs it. However, the *reason* a black body is a perfect radiator is because it absorbs all incident radiation, and by Kirchhoff's law of thermal radiation, a good absorber is also a good emitter.

Why other options are wrong:
- Option A: If both A and R are true, and R is the correct explanation of A - Incorrect because thermal equilibrium doesn't directly explain the radiating property.
- Option C: If A is true but R is false - Incorrect because both A and R are true.
- Option D: If A is false but R is true - Incorrect because A is also true.

Key Concepts:
- Blackbody radiation
- Kirchhoff's law of thermal radiation
- Thermal equilibrium

Memory Aids:
- Remember that a good absorber is a good emitter.

Quick Tips:
- Distinguish between true statements and statements that provide a causal explanation.

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Choose the correct statement among the following

View answer and explanation

Correct option: D

All of the above

Option D (All of the above) is correct because all the statements regarding wave functions, radial distribution, and nodes are accurate.

Solution approach

Correct Answer + Core Reason: All of the above. All the statements are correct definitions and properties related to atomic orbitals and wave functions.

Step-by-Step Solution:

Step 1: Analyze each statement individually.

Step 2: Statement A: ${\mathrm{\Psi }^2}$ represents the atomic orbital. This is correct as ${\mathrm{\Psi }^2}$ gives the probability density of finding an electron in a particular region of space, which defines the shape of the atomic orbital.

Step 3: Statement B: The number of peaks in radial distribution is $n - l$ . This is correct. The number of peaks in the radial distribution function corresponds to the number of radial nodes plus one, which is given by $n - l - 1 + 1 = n - l$ .

Step 4: Statement C: A node is a point in space around the nucleus where the wave function Ψ has zero value. This is the correct definition of a node.

Step 5: Since all the statements are correct, the answer is All of the above.

Why other options are wrong:
- Option A: It only considers the first statement, neglecting the other correct statements.
- Option B: It only considers the second statement, neglecting the other correct statements.
- Option C: It only considers the third statement, neglecting the other correct statements.

Key Concepts:
- Atomic orbitals
- Wave function (Ψ)
- Probability density ( ${\mathrm{\Psi }^2}$ )
- Radial distribution function
- Nodes (radial and angular)

Memory Aids:
- Remember that ${\mathrm{\Psi }^2}$ gives the probability density.
- The number of radial nodes is $n - l - 1$ .
- Total number of nodes is $n-1$ .

Quick Tips:
- Understand the definitions of atomic orbitals, wave functions, and nodes.
- Relate the quantum numbers to the properties of atomic orbitals.

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Application +4 | -1

An electron jumps from an outer orbit to an inner orbit with an energy difference of 3.0eV. What will be the wavelength of the line emitted?

View answer and explanation

Correct option: C

4140 Å

Option C (4140 Å) is correct because it is calculated using the formula λ = hc/E, where E is the energy difference.

Solution approach

Correct Answer + Core Reason: 4140 Å. The wavelength of the emitted photon is calculated using the energy difference and the formula $E = \frac{{hc}}{\lambda}$ .

Step-by-Step Solution:

Step 1: Identify the given information: Energy difference, $E = 3.0 eV$ .\nWe need to find the wavelength of the emitted photon, $\lambda$ .

Step 2: Use the formula $E = \frac{{hc}}{\lambda}$ , where $h = 6.626 \times 10^{-34} Js$ and $c = 3 \times 10^8 m/s$ .

Step 3: Convert the energy from eV to Joules: $E = 3.0 eV \times 1.602 \times 10^{-19} J/eV = 4.806 \times 10^{-19} J$ .

Step 4: Rearrange the formula to solve for $\lambda$ : $\lambda = \frac{{hc}}{E}$

Step 5: Substitute the values: $\lambda = \frac{{(6.626 \times 10^{-34} Js)(3 \times 10^8 m/s)}}{{4.806 \times 10^{-19} J}}$

Step 6: Calculate the wavelength: $\lambda = 4.14 \times 10^{-7} m = 4140 \times 10^{-10} m = 4140 Å$

Step 7: Final Answer: $\lambda = 4140 Å$

Why other options are wrong:
- Option A: This is incorrect. It might be obtained by using an incorrect value for Planck's constant or the speed of light.
- Option B: This is incorrect. It might be obtained by using an incorrect value for Planck's constant or the speed of light.
- Option D: This is incorrect. It might be obtained by using an incorrect value for Planck's constant or the speed of light.

Key Concepts:
- Energy of a photon
- Relationship between energy and wavelength

Memory Aids:
- Remember the formula $E = \frac{{hc}}{\lambda}$

Quick Tips:
- Convert all units to SI units before performing calculations.

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Understanding +4 | -1

The study of photoelectric effect is useful in understanding :

View answer and explanation

Correct option: A

Conservation of energy

Option A (Conservation of energy) is correct because the photoelectric effect demonstrates the conversion of light energy into kinetic energy of electrons.

Solution approach

Correct Answer + Core Reason: Conservation of energy. The photoelectric effect demonstrates the conservation of energy.

Step-by-Step Solution:

Step 1: Describe the photoelectric effect: When a photon strikes a metal surface, it can transfer its energy to an electron in the metal.

Step 2: Explain the energy transfer: If the photon's energy is greater than the work function of the metal, the electron can be ejected from the metal surface with some kinetic energy.

Step 3: State the equation for the photoelectric effect: $E = hf = \phi + KE_{max}$ , where $E$ is the photon energy, $h$ is Planck's constant, $f$ is the frequency of light, $\phi$ is the work function, and $KE_{max}$ is the maximum kinetic energy of the ejected electron.

Step 4: Explain the conservation of energy: The photon's energy is equal to the work function plus the kinetic energy of the electron. This demonstrates the conservation of energy.

Step 5: Final Answer: Conservation of energy

Why other options are wrong:
- Option B: Quantization of charge: The photoelectric effect doesn't directly demonstrate the quantization of charge, although the charge of the electron is quantized.
- Option C: Conservation of charge: The photoelectric effect doesn't directly demonstrate the conservation of charge. Charge is conserved, but the photoelectric effect primarily demonstrates energy conservation.
- Option D: Conservation of kinetic energy: Kinetic energy is not conserved in the photoelectric effect. The photon's energy is converted into work function and kinetic energy of the electron.

Key Concepts:
- Photoelectric effect
- Conservation of energy
- Work function
- Kinetic energy

Memory Aids:
- Photon energy = Work function + Kinetic energy

Quick Tips:
- Remember that the photoelectric effect is a direct application of the conservation of energy.

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Analyzing +4 | -1

Consider the following pairs of electrons

\begin{array}{cccccccccccccccccccc} {\mathrm{(A)}\,\,\mathrm{(a)}\,\mathrm{n = 3,}\,\,\mathrm{l = 1,}\,{\mathrm{m}_\mathrm{l}}\mathrm{ = 1,}\,\,{\mathrm{m}_\mathrm{s}}\mathrm{ = + }\frac{\mathrm{1}}{\mathrm{2}}}\\ {\,\,\,\,\,\,\,\,\,\,\,\mathrm{(b)}\,\mathrm{n = 3,}\,\,\mathrm{l = 2,}\,{\mathrm{m}_\mathrm{l}}\mathrm{ = 1,}\,\,{\mathrm{m}_\mathrm{s}}\mathrm{ = + }\frac{\mathrm{1}}{\mathrm{2}}}\\ {\,\mathrm{(B) (a) n = 3,}\,\,\mathrm{l = 2,}\,{\mathrm{m}_\mathrm{l}}\mathrm{ = } - \mathrm{2,}\,\,{\mathrm{m}_\mathrm{s}}\mathrm{ = } - \frac{\mathrm{1}}{\mathrm{2}}}\\ {\,\,\,\,\,\,\,\mathrm{(b) n = 3,}\,\,\mathrm{l = 2,}\,{\mathrm{m}_\mathrm{l}}\mathrm{ = } - 1\mathrm{,}\,\,{\mathrm{m}_\mathrm{s}}\mathrm{ = } - \frac{\mathrm{1}}{\mathrm{2}}}\\ {\mathrm{(C)}\,\mathrm{(a) n = 4,}\,\,\mathrm{l = 2,}\,{\mathrm{m}_\mathrm{l}}\mathrm{ = 2,}\,\,{\mathrm{m}_\mathrm{s}}\mathrm{ = + }\frac{\mathrm{1}}{\mathrm{2}}}\\ {\,\,\,\,\,\,\,\,\mathrm{(b) n = 3,}\,\,\mathrm{l = 2,}\,{\mathrm{m}_\mathrm{l}}\mathrm{ = 2,}\,\,{\mathrm{m}_\mathrm{s}}\mathrm{ = + }\frac{\mathrm{1}}{\mathrm{2}}} \end{array}

The pairs of electrons present in degenerate orbitals is/are:

View answer and explanation

Correct option: B

Only B

Only B is correct because degenerate orbitals have the same n and l values. In B, both electrons have n=3 and l=2.

Solution approach

Correct Answer + Core Reason: Only B. Degenerate orbitals have the same n and l values.

Step-by-Step Solution:

Step 1: Define degenerate orbitals: Orbitals having the same energy level are called degenerate orbitals. For hydrogen-like atoms, degeneracy depends only on the principal quantum number (n). For multi-electron atoms, degeneracy depends on both n and l.

Step 2: Analyze each pair of electrons:
- Pair A: (a) $n = 3, l = 1$ and (b) $n = 3, l = 2$ . Since $l$ values are different, these orbitals are not degenerate.
- Pair B: (a) $n = 3, l = 2$ and (b) $n = 3, l = 2$ . Since both $n$ and $l$ values are the same, these orbitals are degenerate.
- Pair C: (a) $n = 4, l = 2$ and (b) $n = 3, l = 2$ . Since $n$ values are different, these orbitals are not degenerate.

Step 3: Based on the analysis, only pair B represents electrons in degenerate orbitals.

Why other options are wrong:
- Option A: Pair A has different $l$ values, so they are not degenerate.
- Option C: Pair C has different $n$ values, so they are not degenerate.
- Option D: Only B is correct, C is not.

Key Concepts:
- Degenerate orbitals
- Quantum numbers (n and l)

Memory Aids:
- Remember that for multi-electron atoms, orbitals with the same n and l are degenerate.

Quick Tips:
- Check both n and l values to determine degeneracy in multi-electron atoms.

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Analyzing +4 | -1

Select correct statement given below :

(i) Bohr model is applicable for

{He}^{+}, {Li}^{+}, {Be}^{+}

(ii) Minimum energy required to remove an electron from it's excited state is called as separation energy (for sigle electron species)

(iii) Velocity of electron in a Bohr orbit is

{u}=\frac{2 \pi {KZe}^2}{{nh}}

(iv) Spectrum of hydrogen atom is discontinuous

(v) Paschen series belong from ultraviolet region

View answer and explanation

Correct option: A

(ii), (iii), (iv)

Option A ((ii), (iii), (iv)) is correct because statements (ii), (iii), and (iv) are correct, while (i) is only applicable to single electron species and (v) Paschen series belongs to the infrared region.

Solution approach

Correct Answer + Core Reason: (ii), (iii), (iv). These statements are correct about Bohr's model and related concepts.

Step-by-Step Solution:

Step 1: Analyze statement (i): Bohr model is applicable for ${He}^{+}, {Li}^{+}, {Be}^{+}$ . This statement is incorrect. Bohr's model is applicable only to single-electron species like H, ${He}^{+}$ , ${Li}^{2+}$ , ${Be}^{3+}$ .

Step 2: Analyze statement (ii): Minimum energy required to remove an electron from its excited state is called separation energy (for single electron species). This statement is correct.

Step 3: Analyze statement (iii): Velocity of electron in a Bohr orbit is ${u}=\frac{2 \pi {KZe}^2}{{nh}}$ . This statement is correct.

Step 4: Analyze statement (iv): Spectrum of hydrogen atom is discontinuous. This statement is correct.

Step 5: Analyze statement (v): Paschen series belong from ultraviolet region. This statement is incorrect. Paschen series belongs to the infrared region.

Step 6: Identify the correct statements: (ii), (iii), and (iv).

Why other options are wrong:
- Option B: (i), (iii), (iv): Incorrect because (i) is wrong.
- Option C: (ii), (iii), (v): Incorrect because (v) is wrong.
- Option D: (i), (v): Incorrect because both (i) and (v) are wrong.

Key Concepts:
- Bohr's model
- Single-electron species
- Separation energy
- Velocity of electron in Bohr orbit
- Hydrogen spectrum
- Spectral series (Lyman, Balmer, Paschen, etc.)

Memory Aids:
- Bohr's model is only valid for single-electron species.
- Remember the regions of the electromagnetic spectrum for different spectral series.

Quick Tips:
- Always check if Bohr's model is applicable before applying its formulas.
- Know the regions of the electromagnetic spectrum for different spectral series.

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Understanding +4 | -1

Photoelectric effect shows:

View answer and explanation

Correct option: A

Particle-like behaviour of light

Option A (Particle-like behaviour of light) is correct because the photoelectric effect demonstrates that light can behave as discrete packets of energy (photons), which is a particle-like property.

Solution approach

Correct Answer + Core Reason: Particle-like behaviour of light. Photoelectric effect demonstrates light's particle nature.

Step-by-Step Solution:

Step 1: The question asks what the photoelectric effect demonstrates about the nature of light.

Step 2: Recall that the photoelectric effect is the emission of electrons from a metal surface when light shines on it. This phenomenon can only be explained by considering light as consisting of discrete packets of energy called photons.

Step 3: The energy of a photon is given by $E = hf$ , where h is Planck's constant and f is the frequency of light. This shows that light has particle-like properties.

Why other options are wrong:
- Option B: Wave-like behaviour of light is incorrect because while light also exhibits wave-like behavior (e.g., diffraction, interference), the photoelectric effect specifically demonstrates its particle-like nature.
- Option C: Both wave-like and particle-like behaviour of light is incorrect because the photoelectric effect primarily demonstrates the particle-like behavior.
- Option D: Neither wave-like and particle-like behaviour of light is incorrect because the photoelectric effect demonstrates the particle-like behavior of light.

Key Concepts:
- Photoelectric effect
- Photons
- Particle-wave duality

Memory Aids:
- Photoelectric effect = particles (photons)

Quick Tips:
- Remember that the photoelectric effect is evidence for the particle nature of light.

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Application +4 | -1

Atoms consist of electrons, protons and neutrons. If the mass attributed to neutron was halved and that attributed to the electrons was doubled, the atomic mass of

{\mathrm{\;}_6}{\mathrm{C}^{12}}

would be approximately:

View answer and explanation

Correct option: D

Reduced by 25%

Option D (Reduced by 25%) is correct because the mass of the neutrons is halved, and the mass of the electrons is doubled, resulting in a net decrease in the atomic mass.

Solution approach

Correct Answer + Core Reason: Reduced by 25%. The mass of neutrons is halved and electrons doubled.

Step-by-Step Solution:

Step 1: In ${\mathrm{\;}_6}{\mathrm{C}^{12}}$ , there are 6 protons, 6 neutrons, and 6 electrons.

Step 2: The original atomic mass is approximately the sum of the masses of protons and neutrons (electrons have negligible mass compared to protons and neutrons). So, the original atomic mass is approximately 6(proton mass) + 6(neutron mass).

Step 3: If the mass of the neutron is halved, the new neutron mass is (1/2)(neutron mass). If the mass of the electron is doubled, the new electron mass is 2(electron mass).

Step 4: The new atomic mass would be approximately 6(proton mass) + 6(1/2 neutron mass) + 6(2 electron mass). Since the electron mass is negligible, we can ignore the change in electron mass.

Step 5: New atomic mass ≈ 6(proton mass) + 3(neutron mass).

Step 6: Let's assume the mass of a proton and neutron are approximately equal (m). Then, the original atomic mass is 6m + 6m = 12m. The new atomic mass is 6m + 3m = 9m.

Step 7: The percentage reduction in atomic mass is:
$\frac{{12m - 9m}}{{12m}} \times 100\% = \frac{{3m}}{{12m}} \times 100\% = \frac{1}{4} \times 100\% = 25\%$

Why other options are wrong:
- Option A: Incorrect. The atomic mass changes.
- Option B: Incorrect. The atomic mass decreases.
- Option C: Incorrect. The atomic mass decreases by 25%.

Key Concepts:
- Atomic mass
- Protons, neutrons, and electrons

Memory Aids:
- Atomic mass ≈ mass of protons + mass of neutrons.

Quick Tips:
- Remember that the mass of electrons is negligible compared to protons and neutrons.

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Understanding +4 | -1

The fraction of volume occupied by the nucleus with respect to the total volume of an atom is

View answer and explanation

Correct option: B

{10^{ - 15}}

Option B (

{10^{ - 15}}

) is correct because the nucleus is much smaller than the atom, and the ratio of their volumes is approximately

{10^{ - 15}}

.

Solution approach

Correct Answer + Core Reason: ${10^{ - 15}}$ . The nucleus is much smaller than the atom.

Step-by-Step Solution:

Step 1: Recall the approximate sizes of the atom and the nucleus.
- Radius of an atom: ≈ ${10^{ - 10}}$ m
- Radius of a nucleus: ≈ ${10^{ - 15}}$ m

Step 2: Calculate the volumes of the atom and the nucleus, assuming they are spherical.
- Volume of an atom: $V_{atom} = \frac{4}{3}\pi r_{atom}^3 = \frac{4}{3}\pi ({10^{ - 10}})^3 = \frac{4}{3}\pi \times 10^{ - 30} m^3$
- Volume of a nucleus: $V_{nucleus} = \frac{4}{3}\pi r_{nucleus}^3 = \frac{4}{3}\pi ({10^{ - 15}})^3 = \frac{4}{3}\pi \times 10^{ - 45} m^3$

Step 3: Calculate the fraction of volume occupied by the nucleus with respect to the total volume of the atom:
- $\frac{V_{nucleus}}{V_{atom}} = \frac{\frac{4}{3}\pi \times 10^{ - 45}}{\frac{4}{3}\pi \times 10^{ - 30}} = \frac{10^{ - 45}}{10^{ - 30}} = 10^{ - 15}$

Why other options are wrong:
- Option A: ${10^{ - 16}}$ - Incorrect order of magnitude.
- Option C: ${10^{ - 5}}$ - Incorrect order of magnitude.
- Option D: ${10^{ - 30}}$ - This represents the volume of an atom, not the ratio.

Key Concepts:
- Atomic structure
- Nuclear size
- Volume of a sphere

Memory Aids:
- Remember the approximate sizes of the atom and the nucleus.

Quick Tips:
- Understand the relative sizes of the atom and its nucleus.

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Understanding +4 | -1

The ratio of potential energy and total energy of an electron in a Bohr orbit of Hydrogen like species is:

View answer and explanation

Correct option: A

2

Option A (2) is correct because the potential energy (PE) is -2 times the total energy (TE) in a Bohr orbit, so PE/TE = 2.

Solution approach

Correct Answer + Core Reason: 2. The potential energy is -2 times the total energy in a Bohr orbit.

Step-by-Step Solution:

Step 1: Recall the relationship between potential energy (PE) and total energy (TE) in a Bohr orbit: $TE = -KE$ and $PE = 2TE$ , where KE is kinetic energy.

Step 2: We want to find the ratio $\frac{{PE}}{{TE}}$ .

Step 3: Substitute $PE = 2TE$ into the ratio: $\frac{{PE}}{{TE}} = \frac{{2TE}}{{TE}}$

Step 4: Simplify the expression: $\frac{{PE}}{{TE}} = 2$

Why other options are wrong:
- Option B: -2: This is incorrect because the potential energy is -2 times the total energy, so the ratio is 2, not -2.
- Option C: 1: This is incorrect. The potential energy is not equal to the total energy.
- Option D: -1: This is incorrect. The potential energy is not the negative of the total energy.

Key Concepts:
- Potential energy in Bohr orbit
- Total energy in Bohr orbit
- Kinetic energy in Bohr orbit

Memory Aids:
- Remember: PE = 2TE

Quick Tips:
- Always remember the relationship between potential energy, kinetic energy, and total energy in Bohr's model.

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Analyzing +4 | -1

Which of the following electronic transitions requires for the greatest quantity of energy be absorbed by a H-atom?

View answer and explanation

Correct option: D

n=1

to

n=\infty

Option D (

n=1

to

n=\infty

) is correct because it represents ionization, which requires the greatest amount of energy to completely remove the electron from the atom.

Solution approach

Correct Answer + Core Reason: $n=1$ to $n=\infty$ . This transition requires the greatest energy because it represents ionization.

Step-by-Step Solution:

Step 1: Recall the energy level formula for the hydrogen atom: $E_n = -\frac{13.6}{n^2} \text{ eV}$

Step 2: Calculate the energy difference for each transition:
- A: $n=1$ to $n=2$ : $\Delta E = -13.6(\frac{1}{2^2} - \frac{1}{1^2}) = -13.6(\frac{1}{4} - 1) = 10.2 \text{ eV}$
- B: $n=2$ to $n=4$ : $\Delta E = -13.6(\frac{1}{4^2} - \frac{1}{2^2}) = -13.6(\frac{1}{16} - \frac{1}{4}) = 2.55 \text{ eV}$
- C: $n=3$ to $n=6$ : $\Delta E = -13.6(\frac{1}{6^2} - \frac{1}{3^2}) = -13.6(\frac{1}{36} - \frac{1}{9}) = 0.85 \text{ eV}$
- D: $n=1$ to $n=\infty$ : $\Delta E = -13.6(\frac{1}{\infty^2} - \frac{1}{1^2}) = -13.6(0 - 1) = 13.6 \text{ eV}$

Step 3: Compare the energy differences: The transition $n=1$ to $n=\infty$ has the largest energy difference.

Step 4: Final Answer: $n=1$ to $n=\infty$ requires the greatest quantity of energy.

Why other options are wrong:
- Option A: This transition requires less energy than $n=1$ to $n=\infty$ .
- Option B: This transition requires less energy than $n=1$ to $n=\infty$ .
- Option C: This transition requires less energy than $n=1$ to $n=\infty$ .

Key Concepts:
- Hydrogen atom energy levels
- Electronic transitions
- Ionization energy

Memory Aids:
- Remember that ionization requires the most energy.

Quick Tips:
- Calculate the energy difference for each transition to compare them accurately.
- Remember that ionization corresponds to $n$ approaching infinity.

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Application +4 | -1

The radii of 2nd Bohr orbit of

B{e^{3 + }}

ion is

View answer and explanation

Correct option: B

52.9 pm

Option B (52.9 pm) is correct because the radius of the nth Bohr orbit is given by

r_n = 0.529 \times \frac{n^2}{Z}

Å, where n is the orbit number and Z is the atomic number.

Solution approach

Correct Answer + Core Reason: 52.9 pm. The radius of the nth Bohr orbit is given by $r_n = 0.529 \times \frac{n^2}{Z}$ Å.

Step-by-Step Solution:

Step 1: The radius of the nth Bohr orbit for a hydrogen-like species is given by:
$r_n = 0.529 \times \frac{n^2}{Z} \text{ Å}$ \nwhere n is the principal quantum number (orbit number) and Z is the atomic number.

Step 2: For $Be^{3+}$ , the atomic number (Z) is 4. We are given that n = 2 (2nd Bohr orbit).

Step 3: Substitute the values into the formula:
$r_2 = 0.529 \times \frac{2^2}{4} \text{ Å} = 0.529 \times \frac{4}{4} \text{ Å} = 0.529 \text{ Å}$

Step 4: Convert Å to pm: 1 Å = 100 pm
$r_2 = 0.529 \text{ Å} \times 100 \frac{\text{pm}}{\text{Å}} = 52.9 \text{ pm}$

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

Key Concepts:
- Bohr radius
- Radius of nth Bohr orbit
- Atomic number

Memory Aids:
- $r_n = 0.529 \times \frac{n^2}{Z} \text{ Å}$ (Radius of nth Bohr orbit).

Quick Tips:
- Remember the formula for the radius of the nth Bohr orbit for hydrogen-like species.

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Remembering +4 | -1

For a d-electron, the orbital angular momentum is:

View answer and explanation

Correct option: A

\sqrt{6} \hbar

Option A (

\sqrt{6} \hbar

) is correct because the orbital angular momentum for a d-electron (l=2) is given by

\sqrt{l(l+1)}\hbar

, which equals

\sqrt{6} \hbar

.

Solution approach

Correct Answer + Core Reason: $\sqrt{6} \hbar$ . Orbital angular momentum for a d-electron is calculated using the formula $\sqrt{l(l+1)}\hbar$ .

Step-by-Step Solution:

Step 1: Define the formula for orbital angular momentum: The orbital angular momentum is given by the formula: $L = \sqrt{{l(l+1)}} \hbar$ , where l is the azimuthal quantum number and $\hbar = \frac{h}{{2\pi}}$ .

Step 2: Identify the l value for a d-electron: For a d-electron, l = 2.

Step 3: Substitute l = 2 into the formula: $L = \sqrt{{2(2+1)}} \hbar = \sqrt{{6}} \hbar$

Step 4: Final Answer: The orbital angular momentum for a d-electron is $\sqrt{{6}} \hbar$ .

Why other options are wrong:
- Option B: This option uses $\pi$ instead of $\hbar$ , which is incorrect.
- Option C: This corresponds to l=1 (p-orbital).
- Option D: This option uses $\pi$ instead of $\hbar$ and also has an incorrect value.

Key Concepts:
- Orbital angular momentum
- Azimuthal quantum number (l)

Memory Aids:
- Remember the formula: $L = \sqrt{{l(l+1)}} \hbar$
- s(l=0), p(l=1), d(l=2), f(l=3)... alphabetical order corresponds to increasing l.

Quick Tips:
- Always identify the correct value of l for the given orbital before applying the formula.

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Remembering +4 | -1

If

{r_1}

is the radius of the first orbit, the radius of

{n^{th}}

orbit of H-atom is given by

View answer and explanation

Correct option: A

{r_1}{n^2}

Option A (

{r_1}{n^2}

) is correct because the radius of the nth orbit in a hydrogen atom is proportional to n squared.

Solution approach

Correct Answer + Core Reason: ${r_1}{n^2}$ . The radius of the nth orbit is proportional to n squared.

Step-by-Step Solution:

Step 1: The question asks for the radius of the nth orbit of a hydrogen atom, given the radius of the first orbit, r1.

Step 2: Recall the formula for the radius of the nth orbit in the Bohr model: $r_n = r_1 n^2$ , where $r_1$ is the radius of the first orbit and n is the principal quantum number.

Step 3: Therefore, the radius of the nth orbit is ${r_1}{n^2}$ .

Why other options are wrong:
- Option B: ${r_1}n$ is incorrect because the radius is proportional to $n^2$ , not n.
- Option C: ${r_1}/n$ is incorrect because the radius increases with n, not decreases.
- Option D: $r_1^2{n^2}$ is incorrect because the radius is proportional to $r_1$ , not $r_1^2$ .

Key Concepts:
- Bohr model
- Radius of nth orbit

Memory Aids:
- Radius is proportional to n squared.

Quick Tips:
- Remember the formula for the radius of the nth orbit.

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Understanding +4 | -1

When an electron of charge e and mass m moves with velocity u about the nuclear charge Ze in the circular orbit of radius r, the potential energy of the electron is given by:

View answer and explanation

Correct option: B

- Z{e^2}/r

Option B (

- Z{e^2}/r

) is correct because the potential energy between two charges is negative when they have opposite signs, and is inversely proportional to the distance between them.

Solution approach

Correct Answer + Core Reason: $- Z{e^2}/r$ . The potential energy of an electron in the field of a nucleus is negative and inversely proportional to the distance.

Step-by-Step Solution:

Step 1: The potential energy (PE) between two charges $q_1$ and $q_2$ separated by a distance $r$ is given by: $PE = \frac{{k q_1 q_2}}{{r}}$ , where $k = \frac{1}{{4\pi \epsilon_0}}$

Step 2: In this case, $q_1 = Ze$ (nuclear charge) and $q_2 = -e$ (electron charge).

Step 3: Substituting the charges into the potential energy formula: $PE = \frac{{k (Ze) (-e)}}{{r}} = - \frac{{kZe^2}}{{r}}$

Step 4: Since $k = \frac{1}{{4\pi \epsilon_0}}$ is a constant, we can write the potential energy as $PE = - \frac{{Ze^2}}{{r}}$ (ignoring the constant).

Why other options are wrong:
- Option A: $Z{e^2}/r$ : This is incorrect because it doesn't account for the negative sign arising from the opposite charges of the electron and the nucleus. Potential energy should be negative.
- Option C: $Z{e^2}/{r^2}$ : This is incorrect because the potential energy is inversely proportional to the distance $r$ , not $r^2$ .
- Option D: $m{u^2}/r$ : This represents kinetic energy related to circular motion, not potential energy.

Key Concepts:
- Electrostatic Potential Energy
- Coulomb's Law

Memory Aids:
- Opposite charges attract, leading to negative potential energy.

Quick Tips:
- Remember the sign convention for potential energy based on the charges involved.

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Application +4 | -1

Time period of a wave is

5 \times {10^{ - 3}}s

, what is the frequency?

View answer and explanation

Correct option: B

2 \times {10^2}{\mathrm{s}^{ - 1}}

Option B (

2 \times {10^2}{\mathrm{s}^{ - 1}}

) is correct because frequency is the reciprocal of the time period, and 1 divided by

5 \times 10^{-3}

s equals 200 s

^{-1}

.

Solution approach

Correct Answer + Core Reason: $2 \times {10^2}{\mathrm{s}^{ - 1}}$ . Frequency is the reciprocal of the time period.

Step-by-Step Solution:

Step 1: Recall the relationship between frequency (f) and time period (T): $f = \frac{1}{T}$ .

Step 2: Substitute the given time period: $f = \frac{1}{5 \times 10^{-3} \text{ s}}$ .

Step 3: Calculate the frequency: $f = \frac{10^3}{5} \text{ s}^{-1} = 200 \text{ s}^{-1} = 2 \times 10^2 \text{ s}^{-1}$ .

Why other options are wrong:
- Option A: $5 \times {10^{ - 3}}{\mathrm{s}^{ - 1}}$ - Incorrect, this is the time period, not the frequency.
- Option C: $23 \times {10^3}{\mathrm{s}^{ - 1}}$ - Incorrect, this is not the correct calculation of the reciprocal of the time period.
- Option D: $5 \times {10^2}{\mathrm{s}^{ - 1}}$ - Incorrect, this is not the correct calculation of the reciprocal of the time period.

Key Concepts:
- Frequency
- Time period
- Relationship between frequency and time period

Memory Aids:
- Remember that frequency is the inverse of the time period.

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

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Analyzing +4 | -1

Which of the following transition of electron in the Hydrogen atom will emit maximum energy

View answer and explanation

Correct option: C

{n_3} \to {n_2}

Option C (

{n_3} \to {n_2}

) is correct because the energy difference between

n=3

and

n=2

is the largest among the given options, resulting in the emission of maximum energy.

Solution approach

Correct Answer + Core Reason: ${n_3} \to {n_2}$ . The energy difference is greatest for this transition.

Step-by-Step Solution:

Step 1: The energy emitted during an electronic transition is proportional to the difference in energy levels: $\Delta E = -13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) eV$ , where $n_i$ is the initial energy level and $n_f$ is the final energy level.

Step 2: Calculate the energy difference for each transition:
- A. ${n_5} \to {n_4}$ : $\Delta E_A = -13.6 \left( \frac{1}{4^2} - \frac{1}{5^2} \right) = -13.6 \left( \frac{1}{16} - \frac{1}{25} \right) = -13.6 \left( \frac{9}{400} \right)$
- B. ${n_4} \to {n_3}$ : $\Delta E_B = -13.6 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = -13.6 \left( \frac{1}{9} - \frac{1}{16} \right) = -13.6 \left( \frac{7}{144} \right)$
- C. ${n_3} \to {n_2}$ : $\Delta E_C = -13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = -13.6 \left( \frac{1}{4} - \frac{1}{9} \right) = -13.6 \left( \frac{5}{36} \right)$

Step 3: Compare the energy differences. Since the constant -13.6 is the same for all, we can compare the fractions:
- A: $\frac{9}{400} = 0.0225$
- B: $\frac{7}{144} \approx 0.0486$
- C: $\frac{5}{36} \approx 0.1389$

Step 4: The largest energy difference is for the transition ${n_3} \to {n_2}$ .

Step 5: Final Answer: ${n_3} \to {n_2}$

Why other options are wrong:
- **Option A ( ${n_5} \to {n_4}$ ):** This transition has a smaller energy difference compared to ${n_3} \to {n_2}$ .
- **Option B ( ${n_4} \to {n_3}$ ):** This transition has a smaller energy difference compared to ${n_3} \to {n_2}$ .
- Option D (Same for all): The energy difference is not the same for all transitions.

Key Concepts:
- Hydrogen Atomic Spectra
- Energy Levels
- Electronic Transitions
- Energy Difference and Wavelength

Memory Aids:
- Remember that the energy difference is inversely proportional to the square of the principal quantum number.

Quick Tips:
- The transition with the largest difference in $\frac{1}{n^2}$ will have the maximum energy emitted.

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Analyzing +4 | -1

Given below are two statements.
Statement I According to Bohr’s model of an atom, qualitatively the magnitude of velocity of electron increases with decrease in positive charges on the nucleus as there is no strong hold on the electron by the nucleus.
Statement II According to Bohr’s model of an atom, qualitatively the magnitude of velocity of electron increases with decrease in principal quantum number.
In the light of the above statements, choose the most appropriate answer from the options given below.

View answer and explanation

Correct option: C

Statement I is false but statement II is true.

Statement I is false because increasing positive charge on the nucleus increases the velocity of the electron. Statement II is true because decreasing the principal quantum number increases the velocity of the electron.

Solution approach

Correct Answer + Core Reason: Statement I is false but statement II is true. The velocity of an electron in Bohr's model is directly proportional to the nuclear charge and inversely proportional to the principal quantum number.

Step-by-Step Solution:

Step 1: Recall the formula for the velocity of an electron in the nth orbit according to Bohr's model: $v = \frac{{Z e^2}}{{2 \epsilon_0 n h}}$ , where Z is the atomic number (positive charge on the nucleus), e is the electron charge, $\epsilon_0$ is the permittivity of free space, n is the principal quantum number, and h is Planck's constant.

Step 2: Analyze Statement I: According to Bohr’s model of an atom, qualitatively the magnitude of velocity of electron increases with decrease in positive charges on the nucleus as there is no strong hold on the electron by the nucleus. This statement is false because the velocity is directly proportional to Z. Decreasing Z would decrease the velocity.

Step 3: Analyze Statement II: According to Bohr’s model of an atom, qualitatively the magnitude of velocity of electron increases with decrease in principal quantum number. This statement is true because the velocity is inversely proportional to n. Decreasing n would increase the velocity.

Why other options are wrong:
- Option A: This option is incorrect because Statement II is true.
- Option B: This option is incorrect because Statement I is false.
- Option D: This option is incorrect because Statement II is true.

Key Concepts:
- Bohr's model of the atom
- Velocity of an electron in the nth orbit
- Dependence of velocity on nuclear charge (Z) and principal quantum number (n)

Memory Aids:
- Remember the formula $v = \frac{{Z e^2}}{{2 \epsilon_0 n h}}$

Quick Tips:
- Understand the direct and inverse relationships between velocity, nuclear charge, and principal quantum number.
- Pay attention to the wording of the statements to identify whether they are consistent with Bohr's model.

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Application +4 | -1

The measurement of the electron position is associated with an uncertainty in momentum, which is equal to

1 \times {10^{ - 18}}\;\mathrm{g\;cm\;}{\mathrm{s}^{ - 1}}.

. The uncertainty in electron velocity is: (mass of an electron is

9 \times {10^{ - 28}}\;\mathrm{g}

)

View answer and explanation

Correct option: D

1.1 \times {10^9}\mathrm{cm\;}{\mathrm{s}^{ - 1}}

Option D (

1.1 \times {10^9}\mathrm{cm\;}{\mathrm{s}^{ - 1}}

) is correct because it is calculated using the formula Δv = Δp/m.

Solution approach

Correct Answer + Core Reason: $1.1 \times {10^9}\mathrm{cm\;}{\mathrm{s}^{ - 1}}$ . The uncertainty in electron velocity is calculated using the formula $\Delta v = \frac{{\Delta p}}{m}$ .

Step-by-Step Solution:

Step 1: Identify the given information: Uncertainty in momentum, $\Delta p = 1 \times 10^{-18} g cm/s$ ; mass of electron, $m = 9 \times 10^{-28} g$ .\nWe need to find the uncertainty in electron velocity, $\Delta v$ .

Step 2: Use the formula $\Delta v = \frac{{\Delta p}}{m}$

Step 3: Substitute the values: $\Delta v = \frac{{1 \times 10^{-18} g cm/s}}{{9 \times 10^{-28} g}}$

Step 4: Calculate the uncertainty in velocity: $\Delta v = 1.11 \times 10^9 cm/s \approx 1.1 \times 10^9 cm/s$

Step 5: Final Answer: $\Delta v = 1.1 \times 10^9 cm/s$

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:
- Uncertainty principle
- Relationship between momentum and velocity

Memory Aids:
- Remember the formula $\Delta v = \frac{{\Delta p}}{m}$

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

149 Medium

149 of 193Question permalink

Analyzing +4 | -1

Assertion: All p-orbitals have only one planar node.

Reason: The number of radial nodes depends on the principal quantum number only.

View answer and explanation

Correct option: C

If A is true but R is false

Option C is correct because the assertion is true, but the reason is false. All p-orbitals have one planar node, but the number of radial nodes depends on both n and l.

Solution approach

Correct Answer + Core Reason: If A is true but R is false. The assertion is correct, but the reason is incorrect.

Step-by-Step Solution:

Step 1: Analyze the Assertion (A): All p-orbitals have only one planar node.
- p-orbitals have $l = 1$ . The number of planar nodes is equal to $l$ . Therefore, all p-orbitals have one planar node. The assertion is true.

Step 2: Analyze the Reason (R): The number of radial nodes depends on the principal quantum number only.
- The number of radial nodes is given by the formula: $n - l - 1$ . This formula shows that the number of radial nodes depends on both the principal quantum number (n) and the azimuthal quantum number (l). Therefore, the reason is false.

Step 3: Determine the relationship between A and R.
- Since A is true and R is false, the correct option is C.

Why other options are wrong:
- Option A: R is false, so A cannot be the correct explanation of A.
- Option B: R is false.
- Option D: A is true.

Key Concepts:
- Planar nodes
- Radial nodes
- Quantum numbers (n and l)

Memory Aids:
- Number of planar nodes = l
- Number of radial nodes = n - l - 1

Quick Tips:
- Remember the formulas for calculating the number of planar and radial nodes.

150 Medium

150 of 193Question permalink

Understanding +4 | -1

The threshold energy is given as

{E_0}

and radiation of energy

E

falls on metal, then K.E is given as

View answer and explanation

Correct option: B

E - {E_0}

Option B (

E - {E_0}

) is correct because the kinetic energy (K.E) of an emitted electron is the difference between the energy of the incident radiation (E) and the threshold energy (

{E_0}

), according to the photoelectric effect equation.

Solution approach

Correct Answer + Core Reason: $E - {E_0}$ . Kinetic energy is the difference between incident energy and threshold energy.

Step-by-Step Solution:

Step 1: Recall the photoelectric effect equation: $K.E. = E - {E_0}$ , where K.E. is the kinetic energy of the emitted electron, E is the energy of the incident radiation, and ${E_0}$ is the threshold energy (work function).

Step 2: Apply the equation to the given problem.

Why other options are wrong:
- **Option A ( $\frac{{E - {E_0}}}{2}$ ): This is not the correct equation for kinetic energy in the photoelectric effect.
- Option C ( ${E_0} - E\,$ ): This would result in a negative kinetic energy, which is not physically possible.
- Option D ( $\frac{E}{{{E_0}}}$ ): This is a ratio of energies and does not represent the kinetic energy.

Key Concepts:
- Photoelectric effect
- Kinetic energy
- Threshold energy (work function)

Memory Aids:**
- Remember the photoelectric effect equation: $K.E. = E - {E_0}$ .

Quick Tips:
- Know the photoelectric effect equation and its components.

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