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

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.