Question
In a photoemissive cell, with exciting wavelength $$\lambda ,$$ the fastest electron has speed $$v.$$ If the exciting wavelength is changed to $$\frac{{3\lambda }}{4},$$ the speed of the fastest emitted electron will be
A.
$$v{\left( {\frac{3}{4}} \right)^{\frac{1}{2}}}$$
B.
$$v{\left( {\frac{4}{3}} \right)^{\frac{1}{2}}}$$
C.
less than $$v{\left( {\frac{4}{3}} \right)^{\frac{1}{2}}}$$
D.
greater than $$v{\left( {\frac{4}{3}} \right)^{\frac{1}{2}}}$$
Answer :
greater than $$v{\left( {\frac{4}{3}} \right)^{\frac{1}{2}}}$$
Solution :
Einstein's photoelectric equation is given by
$$KE = E - {W_0}$$
As we know that
$$\eqalign{
& KE = \frac{1}{2}m{v^2}\,\,{\text{and}}\,\,E = \frac{{hc}}{\lambda } \cr
& \therefore \frac{1}{2}m{v^2} = \frac{{hc}}{\lambda } - {W_0}......\left( {\text{i}} \right) \cr} $$
Suppose $${v'}$$ be the new speed, when $$\lambda $$ is changed to $$\frac{{3\lambda }}{4}.$$
The new equation can be written as
$$\frac{1}{2}m{{v'}^2} = \frac{{hc}}{{\left( {\frac{{3\lambda }}{4}} \right)}} - {W_0}$$
$${\text{or}}\,\,\frac{1}{2}m{{v'}^2} = \frac{4}{3}\frac{{hc}}{\lambda } - {W_0}\,......\left( {{\text{ii}}} \right)$$
Dividing Eq. (ii) by Eq. (i), we get
$$\eqalign{
& \frac{{{{v'}^2}}}{{{v^2}}} = \frac{{\frac{4}{3}\frac{{hc}}{\lambda } - {W_0}}}{{\frac{{hc}}{\lambda } - {W_0}}} \cr
& = \frac{{\frac{4}{3}\frac{{hc}}{\lambda } - \frac{4}{3}{W_0} + \frac{1}{3}{W_0}}}{{\frac{{hc}}{\lambda } - {W_0}}} \cr
& = \frac{4}{3} + \frac{{{W_0}}}{{3\left( {\frac{{hc}}{\lambda } - {W_0}} \right)}} > \frac{4}{3} \cr
& \therefore \frac{{v'}}{v} > \sqrt {\frac{4}{3}} \,\,{\text{or}}\,\,v' > \sqrt {\frac{4}{3}v} \cr} $$