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Physics Question on Dual nature of radiation and matter

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.

Answer

The correct answer is: 2.16 eV.
Wavelength of light produced by the argon laser, λ=488nm=488×10−9mλ=488nm=488×10^{−9}m
Stopping potential of the photoelectrons, V0=0.38VV_0=0.38V
1eV=1.6×10−19J1eV=1.6×10^{−19}J
∴V0=0.381.6×10−19eV∴V_0=\frac{0.38}{1.6×10^{-19}}eV
Planck’s constant, h=6.6×10−34Jsh=6.6×10^{−34}Js
Charge on an electron, e=1.6×10−19Ce=1.6×10^{−19}C
Speed of light, c=3×108m/sc=3×10^8m/s
From Einstein’s photoelectric effect, we have the relation involving the work function Φ0Φ_0
of the material of the emitter as:
eV0=hcλ−ϕ0eV_0=\frac{hc}{λ}-ϕ_0
ϕ0=hcλ−eV0ϕ_0=\frac{hc}{λ}-eV_0

=6.6×10−34×3×1081.6×10−19×488×10−9−1.6×10−19×0.381.6×10−19=\frac{6.6×10^{-34}×3×10^8}{1.6×10^{-19}×488×10^{-9}}-\frac{1.6×10^{-19}×0.38}{1.6×10^{-19}}

=2.54−0.38=2.16eV=2.54-0.38=2.16eV
Therefore, the material with which the emitter is made has the work function of 2.16 eV.