Question

Light of wavelength 4000 A˚ is incident on a metal surface. The maximum kinetic energy of emitted photoelectrons is 2 eV. What is the work function of the metal surface?

Answer 1

Step 1. Given data

Wavelength $\left(\lambda \right)$= $4000A^{°}$= $4000\times {10}^{-10}m$

The kinetic energy of electrons = $2eV$

We know, Planks constant, $h$= $6.6\times {10}^{-34}\raisebox{1ex}{$KJ$}\!\left/ \!\raisebox{-1ex}{$Kg-K$}\right.$

Charge on electron = $1.6\times {10}^{-19}C$

Step 2. Calculations

According to the photoelectric equation, the work function of the metal is given by,

$\Phi =\frac{hc}{\lambda }-K{E}_{max}$,

Where $\Phi$ = work function, $h$= planks constant, $c$ = velocity of light, $\lambda$ = wavelength, $K{E}_{max}$= maximum kinetic energy.

$\Phi =\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{4000\times {10}^{-10}\times 1.6\times {10}^{-19}}-2eV=1eV$ (in one Coulomb there are $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\left(1.6\times {10}^{-19}\right)$}\right.$ electrons)

Hence, the work function will be $1eV$.