Question
Photoelectric work function of a metal is $$1\,eV,$$ light of wavelength $$\lambda = 3000\,\mathop {\text{A}}\limits^ \circ $$ falls on it. The photoelectrons come out with velocity
A.
$$10\,m/s$$
B.
$${10^2}\,m/s$$
C.
$${10^4}\,m/s$$
D.
$${10^6}\,m/s$$
Answer :
$${10^6}\,m/s$$
Solution :
According to Einstein's photoelectric equation, $$KE$$ of the photoelectron
\[\frac{1}{2}m{v^2} = \frac{{hc}}{\lambda } - {W_0}\,\,\left[ {\begin{array}{*{20}{c}}
{{\rm{where,}}\,{W_0} = {\rm{work}}\,{\rm{function}}}\\
{\frac{1}{2}m{v^2} = KE\,{\rm{of}}\,{\rm{ejected\,electrons}}}\\
{\frac{{hc}}{\lambda } = {\rm{threshold\,energy}}}
\end{array}} \right]\]
$$\eqalign{
& {\text{Given,}}\,\,\lambda = 3000\,\mathop {\text{A}}\limits^ \circ = 3000 \times {10^{ - 10}}m \cr
& {W_0} = 1\,eV = 1.6 \times {10^{ - 19}}J \cr
& m = 9.1 \times {10^{ - 31}}kg \cr
& \therefore \frac{1}{2} \times 9.1 \times {10^{ - 31}} \times {v^2} \cr
& = \frac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{3000 \times {{10}^{ - 10}}}} - 1.6 \times {10^{ - 19}} \cr
& \Rightarrow v = {10^6}m/s \cr} $$