Question
When a surface is irradiated with light of wavelength $$4950\,\mathop {\text{A}}\limits^ \circ ,$$ a photo current appears which vanishes if a retarding potential greater than $$0.6\,V$$ is applied across the photo tube. When a different source of light is used, it is found that the critical retarding potential is changed to $$1.1\,V.$$ Find the wavelength of second source.
A.
$$4077\,\mathop {\text{A}}\limits^ \circ $$
B.
$$992\,\mathop {\text{A}}\limits^ \circ $$
C.
$$628\,\mathop {\text{A}}\limits^ \circ $$
D.
$$238\,\mathop {\text{A}}\limits^ \circ $$
Answer :
$$4077\,\mathop {\text{A}}\limits^ \circ $$
Solution :
By Einstein equation
$$\eqalign{
& hf = {W_0} + e{V_0}. \cr
& {W_0} = \frac{{hc}}{\lambda } - e{V_0} \cr
& = \frac{{\left( {6.63 \times {{10}^{ - 34}}} \right) \times \left( {3 \times {{10}^8}} \right)}}{{4950 \times {{10}^{ - 10}}}} - \left( {1.6 \times {{10}^{ - 19}}} \right) \times 0.6 \cr
& = 3.12 \times {10^{ - 19}}J \cr} $$
For the second surface
$$\eqalign{
& hf' = {W_0} + e{V_0}^\prime \cr
& {\text{or}}\,\,\frac{{hc}}{{\lambda '}} = 3.12 \times {10^{ - 19}} + 1.6 \times {10^{ - 19}} \times 1.1 \cr
& \therefore \lambda ' = 4077\,\mathop {\text{A}}\limits^ \circ \cr} $$