Class 12 Atoms Mock Test (CBSE)

Topics in Atoms

• Alpha-particle scattering and Rutherford's nuclear model
• Atomic spectra
• Bohr model of the hydrogen atom
• Energy levels
• The line spectra of the hydrogen atom
• de Broglie's explanation of Bohr's second postulate

Atoms practice questions with answers

1. Q1Multiple choice·1 mark
   In the Geiger-Marsden alpha-particle scattering experiment, the number of alpha-particles scattered at an angle $\theta$ is proportional to:

   1. (a)$\frac{1}{\sin(\theta/2)}$
   2. (b)$\frac{1}{\sin^4(\theta/2)}$
   3. (c)$\frac{1}{\sin^3(\theta/2)}$
   4. (d)$\frac{1}{\sin^2(\theta/2)}$
   Show answer and step marks

   1. Correct option: (b) $\frac{1}{\sin^4(\theta/2)}$. The number of scattered $\alpha$-particles $N(\theta)$ varies inversely as $\sin^4(\theta/2)$ as derived from Rutherford's scattering formula.+1

   Answer: (b)
2. Q2Multiple choice·1 mark
   According to de Broglie's explanation of Bohr's second postulate of quantization, the standing wave condition for an electron in an orbit of radius $r$ is:

   1. (a)$\pi r = n\lambda$
   2. (b)$2\pi r = (n+1/2)\lambda$
   3. (c)$2\pi r = n\lambda$
   4. (d)$\pi r = (n+1/2)\lambda$
   Show answer and step marks

   1. Correct option: (c) $2\pi r = n\lambda$. According to de Broglie's hypothesis, the circumference of the Bohr orbit must be equal to an integral multiple of the de Broglie wavelength, leading to $2\pi r = n\lambda$.+1

   Answer: (c)
3. Q3Short answer·3 marks
   The ground state energy of a hydrogen atom is $-13.6 \text{ eV}$. (a) What are the kinetic and potential energies of the electron in this state? (b) Calculate the wavelength of the photon emitted when an electron transitions from the $n = 3$ level to the $n = 1$ level.
   Show answer and step marks

   1. Kinetic energy $K = -E = 13.6 \text{ eV}$.+0.5
   2. Potential energy $U = 2E = -27.2 \text{ eV}$.+0.5
   3. Energy difference for $n = 3 \to n = 1$: $\Delta E = E_3 - E_1 = \frac{-13.6}{3^2} - (-13.6) = 13.6 \times \left(1 - \frac{1}{9}\right) = 12.09 \text{ eV}$.+1
   4. $\lambda = \frac{hc}{\Delta E} = \frac{1242 \text{ eV nm}}{12.09 \text{ eV}} \approx 102.7 \text{ nm}$ (with $hc = 1240$ eV nm, $\approx 102.6$ nm).+1

   Answer: Kinetic energy = $13.6 \text{ eV}$, Potential energy = $-27.2 \text{ eV}$, Wavelength $\approx 102.7 \text{ nm}$ (about $103$ nm)
4. Q4Short answer·3 marks
   Using Bohr's postulates of the atomic model, obtain the expression for the radius of the $n$-th orbit of a hydrogen atom. Hence, find the radius of its first orbit.
   Show answer and step marks

   1. Equate the electrostatic force of attraction between the electron and the nucleus to the required centripetal force: $\frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2} = \frac{m v^2}{r}$+1
   2. Use Bohr's quantization condition for angular momentum: $m v r = \frac{n h}{2\pi}$ and substitute velocity $v$ into the force equation to solve for radius $r_n$.+1
   3. Obtain the expression $r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m e^2}$.+0.5
   4. For $n = 1$: $r_1 = \frac{h^2 \varepsilon_0}{\pi m e^2} \approx 5.3 \times 10^{-11}\text{ m} = 0.53 \text{ \AA}$.+0.5

   Answer: $r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m e^2}$, $r_1 = 0.53 \text{ \AA}$

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