Light of wavelength 200nm shines on an aluminum surface; 4.20 eV is required to eject an electron. What is the kinetic energy of (a) the fastest and (b) the slowest ejected electrons? (c) What is the stopping potential for this situation? (d) What is the cut-off wavelength for aluminum?

The given data is listed below.

The wavelength of light is $\mathrm{\lambda }=200\mathrm{nm}$

The energy required to eject an electron is E = 4.20 eV.

The kinetic energy of a photon is directly proportional to its frequency and inversely proportional to its wavelength. Therefore, kinetic energy on the photon is given by-

$E=\frac{hc}{\lambda }$

The kinetic energy is a scalar, and it is always positive or zero.

The kinetic energy is given by-

$K_{\max }=E_{photon}-\varphi$

Here, $\varphi$is the work function of Aluminium.

The value of is define by,

hc = 1240 eV. nm.

$\begin{array}{c}{K}_{\max }=\frac{hc}{\lambda }-\varphi \\ =\frac{1240eV.nm}{200nm}-4.20eV\\ =2.00eV\end{array}$

Thus, the kinetic energy of the fastest ejected electrons is 2.00 eV.

The kinetic energy of the slowest ejected electrons is zero as the slowest electron will break free of the surface.

The stopping potential $V_0$ is given by-

role="math" localid="1663067795548" $\begin{array}{l}{K}_{\max }=e{V}_0\\ V_0=\frac{K_{\max }}{e}\end{array}$

Substitute 2 eV for $K_{\max }$in the above equation.

$\begin{array}{c}{V}_0=\frac{\left(2.00eV\right)}{e}\\ =2V\end{array}$

Therefore, the stopping potential is 2 V.

For finding cut-off wavelength, take $K_m=0$

$\begin{array}{l}{K}_{\max }=E_{photon}-\varphi \\ 0=\frac{hc}{\lambda }-\varphi \\ \frac{hc}{\lambda }=\varphi \end{array}$

Arrange the above equation for the wavelength as below.

$\begin{array}{l}\frac{hc}{\varphi }=\end{array}\lambda$

Substitute known numerical values in the above equation.

Hence, the cut-off wavelength for Aluminum is 295 nm.