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 $0.38\, V$. 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,
$\lambda=488\, nm =488 \times 10^{-9} m$
Cut-off potential $V_{0}=0.38\, V ,\, e =1.6 \times 10^{-19} C$
Plank constant $h=6.62 \times 10^{-34} J - s$
Velocity of light $c =3 \times 10^{8} m / s$
Let $\phi_{0}$ be the work function.
Use the formula for kinetic energy,
$KE = eV _{0}=\frac{h c}{\lambda}-\phi_{0}$
$1.6 \times 10^{-19} \times 0.38$
$=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{448 \times 10^{-9}}-\phi_{0}$
or $6.08 \times 10^{-20}=40.75 \times 10^{-20}-\phi_{0}$
or $\phi_{0}=(40.75-6.08) \times 10^{-20}=34.67 \times 10^{-20} J$
or $=\frac{34.67 \times 10^{-20}}{1.6 \times 10^{-19}} eV =2.17\, eV$