Question

A surface irradiated with light of wavelength $480\text{nm}$ gives out electrons with maximum velocity $v\text{m/s}$. the cut off wavelength being $600\text{nm}$.The same surface would release electrons with maximum velocity $2v$ m/s if it is irradiated by light of wavelength (in $\text{nm}$) is

Answer 1

From Einstein photo electric equation,
$E=W+(K.E{)}_{max}....(i)$
W = work function $=\frac{hc}{{\lambda }_0}$
$(K.E{)}_{max}=\frac{1}{2}m{v}^2$
E = energy $=\frac{hc}{\lambda }$
$\therefore$ from equation (i),
$(K.E.{)}_{max}=hc[\frac{1}{\lambda }-\frac{1}{{\lambda }_0}]=\frac{1}{2}m{v}^2...(i)$
and $\Rightarrow hc[\frac{1}{{\lambda }^{\prime }}-\frac{1}{{\lambda }_0}]=\frac{1}{2}m(2v{)}^2...(ii)$
equation (i) divided by equation (ii),
$\frac{[\frac{1}{\lambda }-\frac{1}{{\lambda }_0}]}{[\frac{1}{{\lambda }^{\prime }}-\frac{1}{{\lambda }_0}]}=\frac{1}{4}$
$\lambda =480nm,{\lambda }_0=600nm$
now, $\frac{4}{\lambda }-\frac{3}{{\lambda }_0}=\frac{3}{{\lambda }^{\prime }}$
$\frac{4}{480}-\frac{3}{600}=\frac{1}{{\lambda }^{\prime }}$
$\therefore {\lambda }^{\prime }=\frac{200\times 120}{80}=300\text{nm}$