Dual Nature of Radiation and Matter's Question 11 | Physics XII - Part 2 | NCERT Exemplar Questions

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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.

Subjective

(i) In the explanation of photoeletric effect, we assume one photon of frequency $v$ collides with an electron and transfers its energy. This leads to the equation for the maximum energy $E_{\text {max }}$ of the emitted electron as

$$E_{\max }=h v-\phi_0$$

where $\phi_0$ is the work function of the metal. If an electron absorbs 2 photons (each of frequency $v$ ), what will be the maximum energy for the emitted electron?

(ii) Why is this fact (two photon absorption) not taken into consideration in our discussion of the stopping potential?

Explanation

(i) Here it is given that, an electron absorbs 2 photons each of frequency $\nu$ then $\nu^{\prime}=2 v$ where, $\nu^{\prime}$ is the frequency of emitted electron.

Given, $$E_{\max }=h \nu-\phi_0$$

Now, maximum energy for emitted electrons is

$$E_{\max }^{\prime}=h(2 \nu)-\phi_0=2 h \nu-\phi_0$$

(ii) The probability of absorbing 2 photons by the same electron is very low. Hence, such emission will be negligible.

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