Question

The electric field at a point associated with a light wave is
E=(100Vm-1) sin [(3.0×1015 s-1)t] sin [(6.0 ×1015 s-1)t].
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?

Answer 1

The maximum kinetic energy of the photoelectrons K =  hv –  ϕ

Explanation:

Step 1:

Given data in the question

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

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

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

Step 2:

Metal surface work function, $\phi=2 e V$

Highest frequency,

$v=\frac{\omega_{\max }}{2 \pi}=\frac{9 \times 10^{15}}{2 \pi} H z$

From the photoelectric equation of Einstein, Kinetic energy,

$K=h v-\phi$