Question

If a metalic surface is illuminated by radiation of frequency n(n > n0), then the maximum kinetic energy of the emitted photoelectron is K. If the same surface is again illuminated by a radiation of frequency 3n. then the kinetic energy of emitted photoelectrons will be.

Answer 1

Solution

By Einstein's $e{q}^n$ of photoelectric Effect

$hv=h{v}_0+KE$, where

$v\to$ radiation frequency

$v_0\to$ threshold frequency

$KE\to$ maximum kinetic Energy of photoelectric

so, if $v=n$.

$\Rightarrow hn=h{v}_0+KE$

$\Rightarrow hn=h{v}_0+K$ ___ (i)

when $v=3n$

$\Rightarrow h\left(3n\right)=h{v}_0+K{E}^{\prime }$

$\Rightarrow 3hr=h{v}_0+K{E}^{\prime }$ ___ (ii)

Subtracting (ii) from (i), we get

$2hn=K{E}^{\prime }-K$

$\Rightarrow K{E}^{\prime }=2hn+K$