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 $2v\text{m/s}$ if it's irradiated by light of wavelength (in nm)

Answer 1

$\frac{1}{2}m{v}^2=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}.....\left(i\right)$
$\frac{1}{2}m(2v{)}^2=\frac{hc}{{\lambda }^{\prime }}-\frac{hc}{{\lambda }_0}...(ii)$
from equation (1) & (2)
$\frac{4}{\lambda }-\frac{4}{{\lambda }_0}=\frac{1}{{\lambda }^{\prime }}-\frac{1}{{\lambda }_0}$
Now
$\frac{4}{\lambda }-\frac{3}{{\lambda }_0}=\frac{1}{{\lambda }^{\prime }}$
putting ${\lambda }_0=600\text{nm},\lambda =480\text{nm}$
$=\frac{4}{480}-\frac{3}{600}=\frac{1}{{\lambda }^{\prime }}$
$\Rightarrow {\lambda }^{\prime }=\frac{200\times 120}{80}$
we get , ${\lambda }^{\prime }=300\text{nm}$