Question
Radiation of wavelength $$\lambda ,$$ is incident on a photocell. The fastest emitted electron has speed $$v.$$ If the wavelength is changed to $$\frac{{3\lambda }}{4},$$ the speed of the fastest emitted electron will be :
A.
$$ = v{\left( {\frac{4}{3}} \right)^{\frac{1}{2}}}$$
B.
$$ = v{\left( {\frac{3}{4}} \right)^{\frac{1}{2}}}$$
C.
$$ > v{\left( {\frac{4}{3}} \right)^{\frac{1}{2}}}$$
D.
$$ < v{\left( {\frac{4}{3}} \right)^{\frac{1}{2}}}$$
Answer :
$$ > v{\left( {\frac{4}{3}} \right)^{\frac{1}{2}}}$$
Solution :
$$\eqalign{
& hv_0^2 - h{v_0} = \frac{1}{2}m{v^2} \cr
& \therefore \frac{4}{3}h{v_0} - h{v_0} = \frac{1}{2}m{{v'}^2} \cr
& \therefore \frac{{{{v'}^2}}}{{{v^2}}} = \frac{{\frac{4}{3}v - {v_0}}}{{v - {v_0}}} \cr
& \therefore v' = v\sqrt {\frac{{\frac{4}{3}v - {v_0}}}{{v - {v_0}}}} \cr
& \therefore v' > v\sqrt {\frac{4}{3}} \cr} $$