Question

When a beam of 10.6 eV photons of intensity $2.0W{m}^{-2}$ falls on a platinum surface of area $1.0\times {10}^{-4}{m}^2$ and work function 5.6 eV, 0.53% of the incident photons ejects photoelectrons. Find the maximum energy of photoelectron emitted (in eV). Take $1eV=1.6\times {10}^{-19}J$

Answer 1

Given,

Intensity, $I=2W{m}^2$

Area, $A={10}^{-4}{m}^2$

Energy incident on the surface per second,

$P=IA$

$=2.0\times 1.0\times {10}^{-14}$

$=2\times {10}^{-4}J{s}^{-1}$

Energy of each photon,

$=10.6eV$

$=10.6\times 1.6\times {10}^{-19}J$

$\therefore$ Number of photons incident on the surface

$=\frac{2\times {10}^{-4}}{10.6\times 1.6\times {10}^{-19}}$

Number of photoelectrons emitted,

$=\frac{0.53}{100}\times \frac{2\times {10}^{-4}}{10.6\times 1.6\times {10}^{-19}}$

$=6.25\times {10}^{11}$

According to Einstein's photoelectric equation, maximum KE of photoelectrons,

$E_k=ϵ-W=10.6eV-5.6eV$

$=5eV$