Question

Light strikes a sodium surface, causing photoelectric emission. The stopping potential for the ejected electrons is $5.0V$ and the work function of sodium is $2.2eV$. What is the wavelength of the incident light?

Answer 1

The energy of an incident photon is $E=hf$, where $h$ is the Planck constant, and $f$ is the frequency of the electromagnetic radiation. The kinetic energy of the most energetic electron emitted is

$K_m=E-\Phi =\left(hc/\lambda \right)-\Phi$

where $\Phi$ is the work function for sodium, and $f=c/\lambda$, where $\lambda$ is the wavelength of the photon. The stopping potential $V_{\text{stop}}$ is related to the maximum kinetic energy by $e{V}_{\text{stop}}=$

$\begin{array}{c}{K}_m,so\\ e{V}_{stop}=\left(hc/\lambda \right)-\Phi \end{array}$

and

$\lambda =\frac{hc}{e{V}_{stop}+\Phi }=\frac{1240eV\cdot nm}{5.0eV+2.2eV}=170nm$

Here $e{V}_{\text{stop}}=5.0eV$ and $hc=1240eV\cdot nm$ are used.

Note: The cutoff frequency for this problem is

$f_0=\frac{\Phi }{h}=\frac{\left(2.2eV\right)(1.6\times {10}^{-19}J/eV)}{6.626\times {10}^{-34}J\cdot s}=5.3\times {10}^{14}Hz$