Question

The electric field associated with a monochromatic light is given by $E=E_0\sin (1.2\times {10}^{15}\pi t\cdot kx)$. Find the maximum kinetic energy of the photoelectrons when this light falls on a metal surface whose work function is $2.0$ eV.

Answer 1

The electric field becomes$1.2\times {10}^{45}$times per second.

Frequency $=\frac{1.2\times {10}^{15}}{2}=0.6\times {10}^{15}$

$hv={\varphi }_0+KE$

$hv-{\varphi }_0=KE$

$KE=\frac{6.63\times {10}^{-34}\times 0.6\times {10}^{15}}{1.6\times {10}^{-19}}-2=0.48ev$