Question

Light strikes a sodium surface, causing photoelectric emission. The stopping potential for the ejected electrons is 50 V, and the work function of sodium is 2.2 eV. What is the wavelength of the incident light?

Short answer

The wavelength of the incident light is 172nm.

Step-by-step solution

Describe the expression for stopping voltage:

The expression of stopping voltage is given by,

$V_{\mathrm{stop}}=\frac{h}{e}f-\frac{\Phi }{e}$ $V_{stop}=\frac{hc}{\lambda }-\frac{\Phi }{e}$ ….. (1)

Here, h is Planck’s constant, $\lambda$is wavelength, e is charge of electron,$\Phi$is work function, and c is the speed of light.

Determine the wavelength of the incident light:

Consider the given data as below.

The charge,$e=1.6\times {10}^{-19}\mathrm{C}$

The work function of sodium, $\Phi =2.2\mathrm{eV}$

The stopping potential for the ejected electrons, $V_{\mathrm{stop}}=5V$

Plank’s constant, $h=6.626\times {10}^{-34}\mathrm{J}·\mathrm{s}$

Speed of light, $c=3\times {10}^{8}m/s$

put above values in eq 1.

Simplify further for wavelength as below.

$\begin{array}{c}\lambda =\frac{12.423\times {10}^{-7}}{7.2\mathrm{V}}\\ =1.72\times {10}^{-7}\mathrm{m}\\ =172\times {10}^{-9}\mathrm{m}\\ =172\mathrm{nm}\end{array}$

Hence, the wavelength of the incident light is 172 nm.