Question

Find the maximum kinetic energy of the photoelectrons ejected when light of wavelength 350 nm is incident on a cesium surface. Work function of cesium = 1.9 eV.

Answer 1

Here

$\lambda =350nm$

$=350\times {10}^{-5}m,$

$\Phi =1.9eV$

Max. KE of electrons

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

$=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{350\times {10}^{-9}\times 1.6\times {10}^{-19}}-1.9$

$=\frac{6.63\times 3\times {10}^2}{350\times 1.6}-1.9$

$=1.65eV=1.6eV$