Question

In a photo electric effect experiment, the maximum kinetic energy of the emitted electrons is $1eV$ for incoming radiation of frequency $v_0$ and $\text{3 eV}$ for incoming radiation of frequency $\text{3}{\text{V}}_0/2.$ What is the maximum kinetic energy of the electrons emitted for incoming radiations of frequency $\text{9}{\text{V}}_0/4?$

• A. 3 eV
• B. 4.5 eV
• C. 6 eV
• D. 9 eV

Answer 1

Answer: (C)

${(KE)}_{\max }=hv-\text{o}{|}_0$
So, $1eV=h{v}_0-\text{o}{|}_0$ ..(i)
and $3eV=\frac{h{v}_0}{2}-\text{o}{|}_0$ ..(ii)
$\Rightarrow$ $3eV-1eV=\frac{h{v}_0}{2}$ or
$h{v}_0=4eV$
From Eq.(i), $\text{o}{|}_0=h{v}_0-1eV$
$=4eV-1eV$ $=3eV$
$\therefore$ ${(KE)}_{\max }=h\times \frac{9v_0}{4}-3eV$
$=\frac{9}{4}(4eV)-3eV$
$=6eV$