Question

Light of wavelength $488nm$ 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.

• A. 2.2 eV
• B. 3.7 eV
• C. 1.6 eV
• D. 4.2 eV

Answer 1

Answer: (A)

Given, wavelength of light,
$\lambda =488nm=488\times 10^{-9}m$
Cut-off potential $V_0=0.38V,e=1.6\times 10^{-19}C$
Plank constant $h=6.62\times 10^{-34}J-s$
Velocity of light $c=3\times 10^{8}m/s$
Let ${\varphi }_0$ be the work function.
Use the formula for kinetic energy,
$KE=e{V}_0=\frac{hc}{\lambda }-{\varphi }_0$
$1.6\times 10^{-19}\times 0.38$
$=\frac{6.63\times 10^{-34}\times 3\times 10^8}{448\times 10^{-9}}-{\varphi }_0$
or $6.08\times 10^{-20}=40.75\times 10^{-20}-{\varphi }_0$
or ${\varphi }_0=(40.75-6.08)\times 10^{-20}=34.67\times 10^{-20}J$
or $=\frac{34.67\times 10^{-20}}{1.6\times 10^{-19}}eV=2.17eV$