Question

A photon of wavelength $4000\mathring{A}$ strikes a metal surface, the work function of the metal being $2.13eV$. The energy of the photon in $eV$ is:

• A. $E=6.20eV$
• B. $E=1.50eV$
• C. $E=2.50eV$
• D. $E=3.10eV$

Answer 1

The correct option is D $E=3.10eV$

$Energyofphoton=\frac{hc}{wavelength}=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{4000\times {10}^{-10}}J$

$EnergyineV=\frac{EnergyinJoules}{1.6\times {10}^{-14}}=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{1.6\times 4000\times {10}^{-29}}J=3.10eV$