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?
$$\begin{aligned} \text{As,}\quad \lambda & =\frac{h}{\sqrt{2 m q v}} \\ \therefore\quad\lambda & \propto \frac{1}{\sqrt{m q}} \\ \frac{\lambda_p}{\lambda_\alpha} & =\frac{\sqrt{m_\alpha q_\alpha}}{\sqrt{m_p q_p}}=\frac{\sqrt{4 m_p \times 2 e}}{\sqrt{m_p \times e}}=\sqrt{8} \\ \therefore\quad\lambda_p & =\sqrt{8} \lambda_\alpha \end{aligned}$$
i.e., wavelength of proton is $\sqrt8$ times wavelength of $\alpha$-particle.
(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?
(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.
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.
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.
Do all the electrons that absorb a photon come out as photoelectrons?
In photoelectric effect, we can observe that most electrons get scattered into the metal by absorbing a photon.
Thus, all the electrons that absorb a photon doesn't come out as photoelectron. Only a few come out of metal whose energy becomes greater than the work function of metal.
There are two sources of light, each emitting with a power of 100 W . One emits X-rays of wavelength 1 nm and the other visible light at 500 nm . Find the ratio of number of photons of $X$-rays to the photons of visible light of the given wavelength?
Suppose wavelength of $X$-rays is $\lambda_1$ and the wavelength of visible light is $\lambda_2$.
$$\begin{aligned} \text{Given,}\quad P & =100 \mathrm{~W} \\ \lambda_1 & =1 \mathrm{~nm} \\ \text{and}\quad\lambda_2 & =500 \mathrm{~nm} \end{aligned}$$
Also, $n_1$ and $n_2$ represents number of photons of $X$-rays and visible light emitted from the two sources per sec.
So,
$$\begin{aligned} & \frac{E}{t}=P=n_1 \frac{h c}{\lambda_1}=n_2 \frac{h c}{\lambda_2} \\ \Rightarrow \quad & \frac{n_1}{\lambda_1}=\frac{n_2}{\lambda_2}\\ \Rightarrow \quad &\frac{n_1}{n_2}=\frac{\lambda_1}{\lambda_2}=\frac{1}{500} \end{aligned}$$