Question

Photoelectric emission is observed from a surface for frequencies $v_1$ and $v_2$ of the incident radiation $(v_1>v_2)$. If the maximum kinetic energies of the photoelectrons in the two cases are in the ratio $1:k$ then the threshold frequency $v_0$ is given by:

• A. $\frac{v_2-v_1}{k-1}$
• B. $\frac{k{v}_1-v_2}{k-1}$
• C. $\frac{k{v}_2-v_1}{k-1}$
• D. $\frac{v_2-v_1}{k}$

Answer 1

Answer: (B)

$\frac{k{v}_1-v_2}{k-1}$

$h{v}_1=h{v}_0+\frac{1}{2}m{u}_1^2$ ....(1)

$h{v}_2=h{v}_0+\frac{1}{2}m{u}_2^2$ .....(2)

$\frac{1}{2}m{u}_1^2=\left(\frac{1}{k}\right)\frac{1}{2}m{u}_2^2$

From equation (1)-
$h{v}_1=h{v}_0+\frac{1}{2k}m{u}_2^2$ ....(3)

$\frac{1}{2}m{u}_2^2=kh{v}_1-kh{v}_0$.....(4)

From equations (2) and (4), we get-

$h{v}_2=h{v}_0-kh{v}_0+kh{v}_1$

$v_0(1-k)=v_2-k{v}_1$

$v_0=\frac{k{v}_1-v_2}{(k-1)}$