When two monochromatic lights of frequency, $Q$ and $\frac{Q}{2}$ are incident on a photoelectric metal, their

Question

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

Accepted Answer

Using Einstein's photoelectric equation $K_{max} = eV_s = h
u - h
u_0$. For frequency $Q$, $e(V_s/2) = hQ - h
u_0$ (i). For frequency $Q/2$, $eV_s = h(Q/2) - h
u_0$ (ii). From (ii), $h
u_0 = hQ/2 - eV_s$. Substituting into (i): $eV_s/2 = hQ - (hQ/2 - eV_s) \implies eV_s/2 = hQ/2 + eV_s \implies -eV_s/2 = hQ/2$. This implies the values were interchanged. Solving correctly gives $
u_0 = \frac{3}{2} Q$.

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