Question

Light of wavelength 488 𝑛𝑚 is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut - off) potential of photoelectrons is $0.38V$. Find the work function of the material from which the emitter is made.

Answer 1

Formula used: $E=\frac{hc}{\lambda }-W_0=e{V}_0$
Given,
wavelength of light produced by the argon laser, $\lambda =488\text{nm}=488\times {10}^{-9}m$
Stopping potential of the photoelectrons,
$V_0=0.38V$
if the cut-off voltage for the photoelectric emission from the metal is $V_0$ and work function is $W_0$, then the equation for the cut-off energy is given by
$E=\frac{hc}{\lambda }-W_0=e{V}_0$
Here, charge on an electron, $e=1.6\times {10}^{-19}C$
Planck's constant, $h=6.626\times {10}^{-34}Js$
Speed of light, $c=3\times {10}^{8}m/s$
Therefore, the work function for the metal is $W_0=\frac{hc}{\lambda }-e{V}_0$
$W_0=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}\times 488\times {10}^{-9}}-\frac{1.6\times {10}^{-19}\times 0.38}{1.6\times {10}^{-19}}$
$W_0=2.16eV$
Therefore, the material with which the emitter is made has the work function of $2.16eV$
Final answer : $2.16eV$