Question

The electric field at a point associated with a light wave is E = (100 Vm ${}^{-1}$) sin [(3.0 $\times {10}^{15}{s}^{-1}$)t] sin [$(6.0\times {10}^{15}{s}^{-1})t$]. If this light falls on a metal surface having a work function of 2.0 eV, what will be the maximum kinetic energy of the photo electrons ?

Answer 1

Given $E=100\text{sin}\left[\right(3\times {10}^{15}{s}^{-1}\left)t\right]\text{sin}\left[\right(6\times {10}^{-15}{s}^{-1}\left)t\right]$

$=100\frac{1}{2}\left[\text{cos}\right[(9\times {10}^{15}{s}^{-1})t]-\text{cos}[(3\times {10}^{15}{s}^{-1})\left]t\right]$

The W are $9\times {10}^{15}$ and $3\times {10}^{15}$

For largest K.E.

$I_{max}=\frac{W_{max}}{2\pi }=\frac{9\times {10}^{15}}{2\pi }$

$E-{\varphi }_0=\text{K.E.}$

$hf-{\varphi }_0=\text{K.E.}$

$\Rightarrow \frac{6.63\times {10}^{-34}\times 9\times {10}^{15}}{2\pi \times 1.6\times {10}^{-19}}=-2=\text{K.E.}$

$\Rightarrow \text{K.E.}=\frac{6.63\times 9}{2\times 1.6\times 3.14}-2$

$=3.938eV=3.93eV$