Question

The work function of a photoelectric material is 4.0 eV. (a) What is the threshold wavelength ? (b) Find the wavelength of light for which the stopping potential is 2.5 V.

Answer 1

$\Phi =4ev=4\times 1.6\times {10}^{-19}\text{J}$

(a) Threshold wavelength = $\lambda$

$=\frac{hc}{\Phi }$

$=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{4\times 1.6\times {10}^{-19}}$

$=\frac{6.63\times 3}{64}\times \frac{{10}^{-27}}{{10}^{-9}}$

$=3.1\times {10}^{-7}m$

$=310nm$

(b) Stopping potential is 2.5 V

$\text{E}=\Phi +eV$

$\Rightarrow \frac{hc}{\lambda }=4\times 1.6\times {10}^{-19}+1.6\times {10}^{-19}\times 2.5$

$\Rightarrow \lambda =\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{6.5\times 1.6\times {10}^{-19}}$

$=\frac{6.63\times 3\times {10}^{-26}}{1.6\times {10}^{-19}\times 6.2}$

$=1.9125\times {10}^{-7}190nm$