Chapter 12 Atoms NCERT Solutions Class 12 Physics

Question 12.1

Choose the correct alternative from the clues given at the end of each statement:

(a) Atomic Size

The size of the atom in Thomson’s model is no different from the atomic size in Rutherford’s model. (Both assume the atom is roughly $10^{-10}$ m in size; the difference lies in the distribution of charge and mass).

(b) Stability

In the ground state of Thomson’s model, electrons are in stable equilibrium (embedded in positive jelly), while in Rutherford’s model electrons always experience a net force (electrostatic attraction providing centripetal force for orbit).

(c) Collapse

A classical atom based on Rutherford’s model is doomed to collapse. (Accelerated charges radiate energy, causing electrons to spiral into the nucleus).

(d) Mass Distribution

An atom has a nearly continuous mass distribution in a Thomson’s model but has a highly non-uniform mass distribution in Rutherford’s model (mass concentrated in the nucleus).

(e) Positive Charge Mass

The positively charged part of the atom possesses most of the mass in both the models.

Question 12.2

Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect?

Analysis

The key to Rutherford’s scattering experiment is that the target nucleus (Gold) is much heavier than the incident alpha particle.

Mass of Alpha particle ($He^{++}$) $\approx 4u$.
Mass of Hydrogen nucleus (Proton) $\approx 1u$.

Since the target (hydrogen) is lighter than the projectile (alpha particle), the alpha particles will not bounce back (large angle scattering). Instead, the alpha particles will simply push the hydrogen nuclei away and mostly travel forward with slight deviations. Large angle scattering would not be observed.

Question 12.3

A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?

Given Data

Energy Difference $\Delta E = 2.3 \text{ eV}$.
Convert to Joules: $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$.

$$\Delta E = 2.3 \times 1.6 \times 10^{-19} = 3.68 \times 10^{-19} \text{ J}$$

Calculation

Using Planck’s relation $\Delta E = h \nu$:

$$\nu = \frac{\Delta E}{h} = \frac{3.68 \times 10^{-19}}{6.626 \times 10^{-34}}$$ $$\nu = 0.555 \times 10^{15} \approx 5.6 \times 10^{14} \text{ Hz}$$

Frequency $\nu = 5.6 \times 10^{14} \text{ Hz}$.

Question 12.4

The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?

Concept

For an electron in a stable orbit:
Total Energy $E = -13.6 \text{ eV}$.
Kinetic Energy $K = -E$.
Potential Energy $U = 2E$.

Calculation

$$K = -(-13.6) = 13.6 \text{ eV}$$ $$U = 2 \times (-13.6) = -27.2 \text{ eV}$$

Kinetic Energy = 13.6 eV, Potential Energy = -27.2 eV.

Question 12.5

A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.

1. Energy Difference

Energy levels in H-atom: $E_n = \frac{-13.6}{n^2} \text{ eV}$.
Ground state ($n=1$): $E_1 = -13.6 \text{ eV}$.
Excited state ($n=4$): $E_4 = \frac{-13.6}{4^2} = \frac{-13.6}{16} = -0.85 \text{ eV}$.

$$\Delta E = E_4 – E_1 = -0.85 – (-13.6) = 12.75 \text{ eV}$$ $$\Delta E = 12.75 \times 1.6 \times 10^{-19} = 2.04 \times 10^{-18} \text{ J}$$

2. Wavelength and Frequency

$$\nu = \frac{\Delta E}{h} = \frac{2.04 \times 10^{-18}}{6.63 \times 10^{-34}} \approx 3.1 \times 10^{15} \text{ Hz}$$ $$\lambda = \frac{c}{\nu} = \frac{3 \times 10^8}{3.1 \times 10^{15}} \approx 9.7 \times 10^{-8} \text{ m} = 97 \text{ nm}$$

Wavelength = 97 nm, Frequency = $3.1 \times 10^{15} \text{ Hz}$.

Question 12.6

(a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels.

(a) Speed Calculation

Formula: $v_n = \frac{c}{137 n} \approx \frac{2.18 \times 10^6}{n} \text{ m/s}$.

$$v_1 = \frac{2.18 \times 10^6}{1} = 2.18 \times 10^6 \text{ m/s}$$ $$v_2 = \frac{2.18 \times 10^6}{2} = 1.09 \times 10^6 \text{ m/s}$$ $$v_3 = \frac{2.18 \times 10^6}{3} = 7.27 \times 10^5 \text{ m/s}$$

(b) Orbital Period

Formula: $T_n = \frac{2\pi r_n}{v_n}$. Since $r_n \propto n^2$ and $v_n \propto 1/n$, $T_n \propto n^3$.
Using $r_1 = 0.53 \times 10^{-10} \text{ m}$ and $v_1 = 2.18 \times 10^6 \text{ m/s}$:

$$T_1 = \frac{2 \times 3.14 \times 0.53 \times 10^{-10}}{2.18 \times 10^6} \approx 1.52 \times 10^{-16} \text{ s}$$ $$T_2 = 2^3 T_1 = 8 \times 1.52 \times 10^{-16} \approx 1.22 \times 10^{-15} \text{ s}$$ $$T_3 = 3^3 T_1 = 27 \times 1.52 \times 10^{-16} \approx 4.1 \times 10^{-15} \text{ s}$$

Question 12.7

The radius of the innermost electron orbit of a hydrogen atom is $5.3 \times 10^{-11} \text{ m}$. What are the radii of the n = 2 and n = 3 orbits?

Formula

Radius of orbit $r_n = n^2 r_1$.

Calculations

$$r_2 = 2^2 \times r_1 = 4 \times 5.3 \times 10^{-11} = 2.12 \times 10^{-10} \text{ m}$$ $$r_3 = 3^2 \times r_1 = 9 \times 5.3 \times 10^{-11} = 4.77 \times 10^{-10} \text{ m}$$

$r_2 = 2.12 \text{ Å}, \quad r_3 = 4.77 \text{ Å}$

Question 12.8

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

Analysis

Ground state energy $E_1 = -13.6 \text{ eV}$.
Max energy absorbed = 12.5 eV.
Final energy state $E_{final} = -13.6 + 12.5 = -1.1 \text{ eV}$.

Let’s check energy levels:

$n=1 \to E = -13.6$
$n=2 \to E = -3.4$
$n=3 \to E = -1.51$
$n=4 \to E = -0.85$ (Higher than -1.1 eV, so impossible)

The electron can be excited up to the n = 3 state.

Emitted Wavelengths (Transitions)

Possible transitions from $n=3$:

$3 \to 1$ (Lyman Series): $E = -1.51 – (-13.6) = 12.09 \text{ eV}$. $\lambda \approx 102.5 \text{ nm}$.
$3 \to 2$ (Balmer Series): $E = -1.51 – (-3.4) = 1.89 \text{ eV}$. $\lambda \approx 656 \text{ nm}$.
$2 \to 1$ (Lyman Series): $E = -3.4 – (-13.6) = 10.2 \text{ eV}$. $\lambda \approx 121.6 \text{ nm}$.

Lyman Series (2 lines) and Balmer Series (1 line).

Question 12.9

In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius $1.5 \times 10^{11} \text{ m}$ with orbital speed $3 \times 10^4 \text{ m/s}$. (Mass of earth = $6.0 \times 10^{24} \text{ kg}$.)

Formula

Bohr’s Quantization Condition: Angular Momentum $L = mvr = \frac{nh}{2\pi}$.

Calculation

$m = 6.0 \times 10^{24} \text{ kg}$, $v = 3 \times 10^4 \text{ m/s}$, $r = 1.5 \times 10^{11} \text{ m}$.

$$n = \frac{2\pi m v r}{h}$$ $$n = \frac{2 \times 3.14 \times (6.0 \times 10^{24}) \times (3 \times 10^4) \times (1.5 \times 10^{11})}{6.63 \times 10^{-34}}$$ $$n = \frac{169.56 \times 10^{39}}{6.63 \times 10^{-34}}$$ $$n \approx 2.56 \times 10^{74}$$

Quantum Number $n \approx 2.6 \times 10^{74}$. (Very large, indicating classical behavior).