Question

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

(Use h = 6.63 × 10^-34J-s = 4.14 × 10^-15 eV-s, c = 3 × 10⁸ m/s and me = 9.1 × 10^-31kg)

Answer 1

Given :-

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

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

The values of angular frequency ${\displaystyle \omega }$ are ${\displaystyle 9\times {10}^{15}}$ and ${\displaystyle 3\times {10}^{15}}$.

Work function of the metal surface, ${\displaystyle \varphi =2 \text{eV}}$

Maximum frequency,

${\displaystyle v=\frac{{\omega }_{max}}{2\pi }=\frac{9\times {10}^{15}}{2\pi }Hz}$

From Einstein's photoelectric equation, kinetic energy, 

K = hv - ${\displaystyle \varphi }$

${\displaystyle \Rightarrow K=6.63\times {10}^{-34}\times \frac{9\times {10}^{15}}{2\pi }\times \frac{1}{1.6\times {10}^{-19}}-2\text{eV}}$

⇒ K = 3.938 eV