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Atoms Question Bank | Class 12 Physics

CBSE Class 12 › Chapter 12 Notes

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1 Atoms: Question Bank

1.1 Part 1: Multiple Choice Questions (1 Mark)

1.2 Part 2: Assertion-Reason Questions

1.3 Part 3: Important Derivations & Theory

1.4 Part 4: Numericals

1.5 Part 5: Case Study

Atoms: Question Bank

Practice Set: Rutherford Model, Bohr’s Postulates, Hydrogen Spectrum.

10 MCQs 10 Assertion-Reason Case Study

Instructions: Review the Chapter Notes before attempting. Click “Show Answer” to verify.

Part 1: Multiple Choice Questions (1 Mark)

1. In the Rutherford scattering experiment, the number of alpha particles ( $N$ ) scattered at an angle $\theta$ varies as:

(a) $\sin^4(\theta/2)$
(b) $1/\sin^4(\theta/2)$
(c) $\sin^2(\theta/2)$
(d) $1/\sin^2(\theta/2)$

Answer: (b)
The scattering formula derived from the experiment is $N(\theta) \propto \frac{1}{\sin^4(\theta/2)}$ .

2. The radius of the first orbit of hydrogen atom is 0.53 Å. The radius of the second orbit of He $^+$ ion will be:

(a) 1.06 Å
(b) 0.53 Å
(c) 2.12 Å
(d) 0.265 Å

Answer: (a)
Formula: $r_n \propto \frac{n^2}{Z}$ . For He $^+$ , $Z=2$ . For 2nd orbit, $n=2$ .
$r = r_0 \times \frac{2^2}{2} = 0.53 \times 2 = 1.06$ Å.

3. The ground state energy of hydrogen is -13.6 eV. What is the potential energy of the electron in this state?

(a) -13.6 eV
(b) -6.8 eV
(c) -27.2 eV
(d) +13.6 eV

Answer: (c)
Total Energy $E = U/2$ . Thus, $U = 2E = 2 \times (-13.6) = -27.2$ eV.

4. Which spectral series of the hydrogen spectrum lies in the visible region?

(a) Lyman
(b) Balmer
(c) Paschen
(d) Pfund

Answer: (b)
Balmer series transitions ( $n_f = 2$ ) emit photons in the visible range (H-alpha, H-beta, etc.).

5. According to Bohr’s quantization condition, the angular momentum of an electron in the 3rd orbit is:

(a) $h/\pi$
(b) $3h/2\pi$
(c) $3h/\pi$
(d) $h/2\pi$

Answer: (b)
$L = \frac{nh}{2\pi}$ . For $n=3$ , $L = \frac{3h}{2\pi} = 1.5 \frac{h}{\pi}$ .

6. The distance of closest approach of an alpha particle projected towards a nucleus depends on kinetic energy $K$ as:

(a) $K$
(b) $K^2$
(c) $1/K$
(d) $1/K^2$

Answer: (c)
At distance of closest approach ( $r_0$ ), Kinetic Energy = Potential Energy $\propto 1/r_0$ . Thus, $r_0 \propto 1/K$ .

7. The ratio of maximum to minimum wavelength in the Lyman series is:

(a) 4:3
(b) 9:8
(c) 3:4
(d) 1:1

Answer: (a)
$\lambda_{max}$ (transition $2 \to 1$ ): $1/\lambda = R(1 - 1/4) = 3R/4 \Rightarrow \lambda = 4/3R$ .
$\lambda_{min}$ (transition $\infty \to 1$ ): $1/\lambda = R(1 - 0) = R \Rightarrow \lambda = 1/R$ .
Ratio = $(4/3R) / (1/R) = 4:3$ .

8. What is the ionization energy of a Hydrogen atom in the first excited state ( $n=2$ )?

(a) 13.6 eV
(b) 3.4 eV
(c) 10.2 eV
(d) 1.51 eV

Answer: (b)
Energy in $n=2$ is $E_2 = -13.6 / 2^2 = -3.4$ eV. Energy required to remove it ( $0 - E_2$ ) is $+3.4$ eV.

9. When an electron jumps from $n=4$ to $n=1$ , the number of spectral lines possible is:

(a) 3
(b) 4
(c) 6
(d) 10

Answer: (c)
Number of lines = $n(n-1)/2$ . For $n=4$ , $4(3)/2 = 6$ lines.

10. Bohr’s model is valid for:

(a) Hydrogen atom only
(b) All atoms
(c) Single electron species (H, He $^+$ , Li $^{++}$ )
(d) Molecules

Answer: (c)
Bohr’s model relies on the interaction between a nucleus and a single electron, ignoring electron-electron interactions.

Part 2: Assertion-Reason Questions

(A) Both A & R are true, R explains A.
(B) Both A & R are true, R does NOT explain A.
(C) A is true, R is false.
(D) A is false, R is true.

1. Assertion (A): An atom is electrically neutral.
Reason (R): Atoms contain equal number of protons and neutrons.

Answer: (C)
Assertion is True. Reason is False. Neutrality is due to equal protons and electrons, not neutrons.

2. Assertion (A): The total energy of an electron in a hydrogen atom is negative.
Reason (R): The electron is bound to the nucleus by attractive electrostatic force.

Answer: (A)
Correct. Negative energy signifies a bound system; energy must be supplied to free the electron.

3. Assertion (A): In the Rutherford experiment, $\alpha$ -particles were chosen.
Reason (R): $\alpha$ -particles have a high mass and momentum, suffering small deflections by electrons.

Answer: (A)
Correct. If lighter particles were used, they would be easily deflected by orbiting electrons, obscuring the effect of the nucleus.

4. Assertion (A): An electron does not radiate energy while moving in a stationary orbit.
Reason (R): The angular momentum of the electron in a stationary orbit is quantized ( $nh/2\pi$ ).

Answer: (A)
Correct. This is Bohr’s postulate. The quantization condition defines the stability of these specific orbits.

5. Assertion (A): Balmer series lies in the visible region of the spectrum.
Reason (R): For Balmer series, $n_f = 1$ .

Answer: (C)
Assertion is True. Reason is False. For Balmer series, the final state is $n_f = 2$ .

6. Assertion (A): Large angle scattering of alpha particles led to the discovery of the nucleus.
Reason (R): The entire positive charge and mass of the atom is concentrated in a small volume.

Answer: (A)
Correct. Only a dense concentration of positive charge could repel a heavy, energetic $\alpha$ -particle back by $>90^\circ$ .

7. Assertion (A): The velocity of an electron increases as it moves to higher orbits.
Reason (R): Velocity is inversely proportional to the principal quantum number $n$ .

Answer: (D)
Assertion is False. Since $v \propto 1/n$ , velocity decreases in higher orbits. Reason is True.

8. Assertion (A): Hydrogen atom has only one electron but its spectrum has many lines.
Reason (R): A sample of hydrogen contains millions of atoms, and electrons in different atoms transition between different energy levels.

Answer: (A)
Correct. While a single atom emits one photon at a time, the bulk gas emits all possible transitions simultaneously.

9. Assertion (A): Electrons in atoms can only occupy certain discrete energy levels.
Reason (R): De Broglie waves of the electron interfere destructively if the orbit circumference is not an integral multiple of wavelength.

Answer: (A)
Correct. This explains the physical basis of Bohr’s quantization condition ( $2\pi r = n\lambda$ ).

10. Assertion (A): The impact parameter is zero for a head-on collision.
Reason (R): In a head-on collision, the scattering angle is $180^\circ$ .

Answer: (B)
Both are true. Impact parameter $b=0$ implies the particle is aimed directly at the nucleus center, leading to a rebound ( $180^\circ$ ). However, R is a consequence of A, not the explanation of the definition of impact parameter.

Part 3: Important Derivations & Theory

1. Using Bohr’s postulates, derive the expression for the radius of the $n^{th}$ orbit of a hydrogen atom.

Step 1: Centripetal force = Electrostatic force.
$\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0}\frac{e^2}{r^2} \Rightarrow mv^2 = \frac{e^2}{4\pi\epsilon_0 r}$ … (i)
Step 2: Quantization condition.
$mvr = \frac{nh}{2\pi} \Rightarrow v = \frac{nh}{2\pi mr}$ … (ii)
Step 3: Substitute (ii) in (i):
$m \left(\frac{nh}{2\pi mr}\right)^2 = \frac{e^2}{4\pi\epsilon_0 r}$
Solving for $r$ : $r_n = \frac{\epsilon_0 n^2 h^2}{\pi m e^2}$ .

2. State Bohr’s three postulates of the atomic model.

1. Stationary Orbits: Electrons revolve around the nucleus in specific stable orbits without radiating energy.
2. Quantization Condition: The electron revolves only in those orbits where its angular momentum is an integral multiple of $h/2\pi$ . ( $L = \frac{nh}{2\pi}$ ).
3. Frequency Condition: Energy is emitted or absorbed only when an electron jumps from one orbit to another. ( $h\nu = E_i - E_f$ ).

3. Draw the energy level diagram of Hydrogen and mark the Lyman, Balmer, and Paschen series.

Lyman: Transitions to $n=1$ (UV region).
Balmer: Transitions to $n=2$ (Visible region).
Paschen: Transitions to $n=3$ (Infrared region).

Part 4: Numericals

1. A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Ground state energy $E_1 = -13.6$ eV.
Energy supplied = 12.5 eV. Max Energy = $-13.6 + 12.5 = -1.1$ eV.
Let’s check possible levels:
$E_1 = -13.6$ eV
$E_2 = -3.4$ eV
$E_3 = -1.51$ eV
$E_4 = -0.85$ eV
Since $-1.1$ eV is between $E_3$ and $E_4$ , the electron can go up to $n=3$ .
Possible transitions: $3\to1$ (Lyman), $3\to2$ (Balmer), $2\to1$ (Lyman).
Answer: Lyman and Balmer series lines.

2. Calculate the shortest wavelength present in the Paschen series of spectral lines. (Given $R = 1.097 \times 10^7$ m $^{-1}$ ).

For Paschen series, $n_f = 3$ .
Shortest wavelength ( $\lambda_{min}$ ) corresponds to max energy transition: $n_i = \infty \to n_f = 3$ .
$\frac{1}{\lambda} = R \left(\frac{1}{3^2} - \frac{1}{\infty^2}\right) = \frac{R}{9}$ .
$\lambda = \frac{9}{R} = \frac{9}{1.097 \times 10^7} \approx 8.2 \times 10^{-7}$ m.
$\lambda \approx 820$ nm.

3. In a Geiger-Marsden experiment, what is the distance of closest approach to the nucleus of a 7.7 MeV $\alpha$ -particle before it comes momentarily to rest and reverses its direction? (Z for Gold = 79).

At closest approach, Kinetic Energy = Potential Energy.
$K = \frac{1}{4\pi\epsilon_0} \frac{(2e)(Ze)}{r_0}$
$r_0 = \frac{2Ze^2}{4\pi\epsilon_0 K}$
$K = 7.7 \text{ MeV} = 7.7 \times 1.6 \times 10^{-13}$ J.
$r_0 = \frac{9 \times 10^9 \times 2 \times 79 \times (1.6 \times 10^{-19})^2}{7.7 \times 1.6 \times 10^{-13}}$
$r_0 \approx 30$ fm ( $3.0 \times 10^{-14}$ m).

Part 5: Case Study

Case Study: The Hydrogen Spectrum
The atomic hydrogen spectrum consists of several series of lines. These lines correspond to the radiation emitted when an electron jumps from a higher energy state to a lower energy state. The wavelengths are given by the Rydberg formula. This discrete spectrum was the first strong evidence that energy levels in atoms are quantized. The study of these lines allows astronomers to determine the composition of distant stars.

Why is the spectrum of hydrogen a line spectrum and not a continuous one?
Which series lies in the ultraviolet region?
Find the ratio of the longest to shortest wavelength in the Balmer series.
What happens to the spacing between energy levels as $n$ increases?

1. Because electrons can only occupy specific, quantized energy levels. Light is emitted only at specific frequencies corresponding to the energy difference between these levels.
2. Lyman Series. ( $n_f = 1$ )
3. Calculation:
Longest ( $\lambda_L$ ): $3 \to 2$ . $1/\lambda_L = R(1/4 - 1/9) = 5R/36$ .
Shortest ( $\lambda_S$ ): $\infty \to 2$ . $1/\lambda_S = R(1/4 - 0) = R/4$ .
Ratio $\frac{\lambda_L}{\lambda_S} = \frac{36/5R}{4/R} = \frac{36}{20} = \textbf{9:5}$ .
4. The energy levels get closer together (crowded) as $n$ increases, because $E \propto 1/n^2$ .