# Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect.

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.

Solution:

Wavelength of light produced by the argon laser, λ = 488 nm

= 488 × 10−9 m

Stopping potential of the photoelectrons, V0 = 0.38 V

1eV = 1.6 × 10−19 J

$\therefore V_{0}=\frac{0.38}{1.6 \times 10^{-19}} \mathrm{eV}$

Planck's constant, $h=6.6 \times 10^{-34} \mathrm{Js}$

Charge on an electron, e = 1.6 × 10−19 C

Speed of light, c = 3 × 10 m/s

From Einstein’s photoelectric effect, we have the relation involving the work function Φ0 of the material of the emitter as:

$e V_{0}=\frac{h c}{\lambda}-\phi_{0}$

$\phi_{0}=\frac{h c}{\lambda}-e V_{0}$

$=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{1.6 \times 10^{-19} \times 488 \times 10^{-9}}-\frac{1.6 \times 10^{-19} \times 0.38}{1.6 \times 10^{-19}}$

$=2.54-0.38=2.16 \mathrm{eV}$

Therefore, the material with which the emitter is made has the work function of 2.16 eV.