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12
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

A
decreases with increasing $n$, with $v$ fixed
B
decreases with $n$ fixed, $v$ increasing
C
remains constant with $n$ and $v$ changing such that $n v=$ constant
D
increases when the product $n v$ increases
13
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?

A
The particle could be moving in a circular orbit with origin as centre
B
The particle could be moving in an elliptic orbit with origin as its focus
C
When the de-Broglie wavelength is $\lambda_1$, the particle is nearer the origin than when its value is $\lambda_2$
D
When the de-Broglie wavelength is $\lambda_2$, the particle is nearer the origin than when its value is $\lambda_1$
14
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.

15
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.

16
Subjective

There are materials which absorb photons of shorter wavelength and emit photons of longer wavelength. Can there be stable substances which absorb photons of larger wavelength and emit light of shorter wavelength.

Explanation

According to first statement, when the materials which absorb photons of shorter wavelength has the energy of the incident photon on the material is high and the energy of emitted photon is low when it has a longer wavelength. But in second statement, the energy of the incident photon is low for the substances which has to absorb photons of larger wavelength and energy of emitted photon is high to emit light of shorter wavelength. This means in this statement material has to supply the energy for the emission of photons.

But this is not possible for a stable substances.