Question

Photo electric emission is observed from a surface for frequencies $v_1$ and $v_2$. The Kinetic Energy in two cases are in ratio 1 : K, then the threshold frequency $v_o$ is given by:

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

Answer 1

The correct option is B $\frac{K{v}_1-v_2}{K-1}$

Let,

$h{v}_1=h{v}_o+1⟹h[v_1-v_o]=1$ ------- (1)

$h{v}_2=h{v}_o+K⟹h[v_2-v_o]=K$ -------- (2)

Dividing equation 1 and 2,

$\frac{1}{K}=\frac{v_1-v_o}{v_2-v_o}$

$v_2-v_o=K{v}_1-K{v}_o$

$K{v}_o-v_o=K{v}_1-v_2$

$v_o[K-1]=K{v}_1-v_2$

$v_o=\frac{K{v}_1-v_2}{K-1}$

Option B is correct.