Question

When two monochromatic light of frequency ν and $\frac{ν}{2}$ are incident on a photoelectric metal, their stopping potential becomes $\frac{V_s}{2}$ and Vs respectively. The threshold frequency for this metal is:

• A. 2 v
• B. 3 v
• C. $\frac{2}{3v}$
• D. $\frac{3}{2v}$

Answer 1

Answer: (D)

$\frac{3}{2v}$

Answer 2

$hv = W + 2v_0 e$
$2hv = W + \frac{v_0}{e}$
on solving we get,$W = \frac{3}{2}hv$
$h_0 = \frac{3}{2}hv$
$v_0 = \frac{3}{2}v$