Question

Photo-electric 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 photo-electrons 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. $\frac{K{v}_2-v_1}{K-1}$
• C. $\frac{K{v}_1-v_2}{K-1}$
• D. $\frac{v_1-v_2}{K}$

Answer 1

The correct option is B $\frac{K{v}_2-v_1}{K-1}$

As we know,
$hv=h{v}_0+\left(1/2\right)m{u}^2$
So according to given conditions.
$h{v}_1=h{v}_0+\frac{1}{2}m{u}_1^2$ .......(i)
$h{v}_2=h{v}_0+\frac{1}{2}m{u}_2^2$ ........(ii)
$\because \frac{1}{2}m{u}_1^2=\frac{1}{k}\left(\frac{1}{2}m{u}_2^2\right)$
$\therefore$ from Eq.(i) $h{v}_1=h{v}_0+\frac{1}{2k}m{u}_2^2$ ......(iii)
or $\frac{1}{2}m{u}_2^2=kh{v}_1-kh{v}_0$ ........ (iv)
By, Eqs. (ii) and (iv),
$h{v}_2=h{v}_0-kh{v}_0+kh{v}_1$
or $v_0(1-k)=v_2-k{v}_1$
or $v_0=\frac{k{v}_1-v_2}{(k-1)}$

Hence, the correct option is $B$