Question

Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light rays ($v_1>v_2$) . If the maximum values of kinetic energy of the photo-electrons emitted in the two cases are in ratio of 1:k, then the threshold frequency of the metallic surface is:

• A. $\frac{v_{1}k-v_2}{k-1}$
• B. $\frac{v_{2}k-v_1}{k-1}$
• C. $\frac{v_{1}k-v_2}{1-k}$
• D. $\frac{v_1-v_{2}k}{k-1}$

Answer 1

The correct option is A $\frac{v_{1}k-v_2}{k-1}$

Maximum Kinetic Energy, $K.E{.}_{max}=hv-h{v}_T$

$K.E{.}_{max\left(1\right)}=h{v}_1-h{v}_T.....1$

$K.E{.}_{max\left(2\right)}=h{v}_2-h{v}_T.....2$

Dividing 1 by 2:

$\frac{K.E{.}_{max\left(1\right)}}{K.E{.}_{max\left(2\right)}}=\frac{h{v}_1-h{v}_T}{h{v}_2-h{v}_T}$

$\frac{1}{k}=\frac{h{v}_1-h{v}_T}{h{v}_2-h{v}_T}$

$h{v}_2-h{v}_T=h{v}_{1}k-h{v}_{T}k$

$h{v}_T(k-1)=h{v}_{1}k-h{v}_2$

$v_T=\frac{v_{1}k-v_2}{k-1}$

So, the answer is (A).