Dual Nature of Radiation and Matter's Question 10 | Physics XII - Part 2 | NCERT Exemplar Questions

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MCQ (Multiple Correct Answer)

Two particles $A_1$ and $A_2$ of masses $m_1, m_2\left(m_1>m_2\right)$ have the same de-Broglie wavelength. Then,

their momenta are the same

their energies are the same

energy of $A_1$ is less than the energy of $A_2$

energy of $A_1$ is more than the energy of $A_2$

MCQ (Multiple Correct Answer)

The de-Broglie wavelength of a photon is twice, the de-Broglie wavelength of an electron. The speed of the electron is $v_e=\frac{c}{100}$. Then,

$\frac{E_e}{E_p}=10^{-4}$

$\frac{E_e}{E_p}=10^{-2}$

$\frac{p_e}{m_e c}=10^{-2}$

$\frac{p_e}{m_e C}=10^{-4}$

MCQ (Multiple Correct Answer)

Photons absorbed in matter are converted to heat. A source emitting $n$ photon $/ \mathrm{sec}$ of frequency $v$ is used to convert 1 kg of ice at $0^{\circ} \mathrm{C}$ to water at $0^{\circ} \mathrm{C}$. Then, the time $T$ taken for the conversion

decreases with increasing $n$, with $v$ fixed

decreases with $n$ fixed, $v$ increasing

remains constant with $n$ and $v$ changing such that $n v=$ constant

increases when the product $n v$ increases

MCQ (Multiple Correct Answer)

A particle moves in a closed orbit around the origin, due to a force which is directed towards the origin. The de-Broglie wavelength of the particle varies cyclically between two values $\lambda_1, \lambda_2$ with $\lambda_1>\lambda_2$. Which of the following statement are true?

The particle could be moving in a circular orbit with origin as centre

The particle could be moving in an elliptic orbit with origin as its focus

When the de-Broglie wavelength is $\lambda_1$, the particle is nearer the origin than when its value is $\lambda_2$

When the de-Broglie wavelength is $\lambda_2$, the particle is nearer the origin than when its value is $\lambda_1$

Subjective

A proton and an $\alpha$-particle are accelerated, using the same potential difference. How are the de-Broglie wavelengths $\lambda_p$ and $\lambda_\alpha$ related to each other?

Explanation

$$\begin{aligned} \text{As,}\quad \lambda & =\frac{h}{\sqrt{2 m q v}} \\ \therefore\quad\lambda & \propto \frac{1}{\sqrt{m q}} \\ \frac{\lambda_p}{\lambda_\alpha} & =\frac{\sqrt{m_\alpha q_\alpha}}{\sqrt{m_p q_p}}=\frac{\sqrt{4 m_p \times 2 e}}{\sqrt{m_p \times e}}=\sqrt{8} \\ \therefore\quad\lambda_p & =\sqrt{8} \lambda_\alpha \end{aligned}$$

i.e., wavelength of proton is $\sqrt8$ times wavelength of $\alpha$-particle.

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