When two monochromatic lights of frequency, $\nu$ and $\dfrac{\nu}{2}$ are incident on a photoelectric metal, their stopping potential becomes $\dfrac{V_s}{2}$ and $V_s$ respectively. The threshold frequency for this metal is
Answer: (A) $\dfrac{3}{2}\nu$
Einstein's equation $eV = h\nu - h\nu_0$ for the two cases:
$$\frac{eV_s}{2} = h\nu - h\nu_0 \qquad (1)$$
$$eV_s = \frac{h\nu}{2} - h\nu_0 \qquad (2)$$
Doubling (1) and equating with (2):
$$2h\nu - 2h\nu_0 = \frac{h\nu}{2} - h\nu_0 \;\Rightarrow\; h\nu_0 = \frac{3}{2}h\nu \;\Rightarrow\; \nu_0 = \frac{3}{2}\nu$$
Note: this threshold is higher than both incident frequencies, so with these data no photoemission could actually occur. The data are not physically consistent, but solving the two equations as given leads to $\dfrac{3}{2}\nu$, the answer expected in the exam.
Solution by Sreeraj P, M.Sc Physics