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,
$$\begin{aligned} & \frac{E}{t}=P=n_1 \frac{h c}{\lambda_1}=n_2 \frac{h c}{\lambda_2} \\ \Rightarrow \quad & \frac{n_1}{\lambda_1}=\frac{n_2}{\lambda_2}\\ \Rightarrow \quad &\frac{n_1}{n_2}=\frac{\lambda_1}{\lambda_2}=\frac{1}{500} \end{aligned}$$
Consider figure for photoemission. How would you reconcile with momentum-conservation? Note light (photons) have momentum in a different direction than the emitted electrons.
During photoelectric emission, the momentum of incident photon is transferred to the metal. At microscopic level, atoms of a metal absorb the photon and its momentum is transferred mainly to the nucleus and electrons.
The excited electron is emitted. Therefore, the conservation of momentum is to be considered as the momentum of incident photon transferred to the nucleus and electrons.
Consider a metal exposed to light of wavelength 600 nm . The maximum energy of the electron doubles when light of wavelength 400 nm is used. Find the work function in eV.
Given,
For the first condition, Wavelength of light $\lambda=600 \mathrm{~nm}$ and for the second condition, Wavelength of light $\lambda^{\prime}=400 \mathrm{~nm}$
Also, maximum kinetic energy for the second condition is equal to the twice of the kinetic energy in first condition.
i.e., $$K_{\max }^{\prime}=2 K_{\max }$$
$$\begin{aligned} & \begin{aligned} \text { Here, } \quad K_{\max }^{\prime} & =\frac{h c}{\lambda}-\phi \\ \Rightarrow \quad 2 K_{\max } =\frac{h c}{\lambda^{\prime}}-\phi_0 \\ \Rightarrow \quad 2\left(\frac{1230}{600}-\phi\right) & =\left(\frac{1230}{400}-\phi\right) \quad[\because h c \approx 1240 \mathrm{eVnm}] \\ \Rightarrow \quad \phi =\frac{1230}{1200}=1.02 \mathrm{eV} \end{aligned} \end{aligned}$$
Assuming an electron is confined to a 1 nm wide region, find the uncertainty in momentum using Heisenberg uncertainty principle $(\Delta x \times \Delta p \approx h)$. You can assume the uncertainty in position $\Delta x$ as 1 nm . Assuming $p \approx \Delta p$, find the energy of the electron in electronvolts.
Here, $\Delta x=1 \mathrm{~nm}=10^{-9} \mathrm{~m}, \Delta p=$ ?
As $\Delta x \Delta p \approx h$
$$\therefore\quad\Delta p=\frac{h}{\Delta x}=\frac{h}{2 \pi \Delta x}$$
$$\begin{aligned} \Rightarrow \quad & =\frac{6.62 \times 10^{-34} \mathrm{Js}}{2 \times(22 / 7)\left(10^{-9}\right) \mathrm{m}} \\ & =1.05 \times 10^{-25} \mathrm{~kg} \mathrm{~m} / \mathrm{s} \end{aligned}$$
$$\begin{aligned} &\text { Energy, }\quad E=\frac{p^2}{2 m}=\frac{(\Delta p)^2}{2 m} \quad[\because p \approx \Delta p] \end{aligned}$$
$$ \begin{aligned} & =\frac{\left(1.05 \times 10^{-25}\right)^2}{2 \times 9.1 \times 10^{-31}} \mathrm{~J} \\ \Rightarrow\quad & =\frac{\left(1.05 \times 10^{-25}\right)^2}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19}} \mathrm{eV} \\ & =3.8 \times 10^{-2} \mathrm{eV} \end{aligned}$$