Solved: PHYSICS Question for NEET

NEET PHYSICSHard

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

$2 Q$

$3 Q$

$\frac{2}{3} Q$

$\frac{3}{2} Q$

Step-by-Step Solution

Using Einstein's photoelectric equation $K_{max} = eV_s = h\nu - h\nu_0$ . For frequency $Q$ , $e(V_s/2) = hQ - h\nu_0$ (i). For frequency $Q/2$ , $eV_s = h(Q/2) - h\nu_0$ (ii). From (ii), $h\nu_0 = hQ/2 - eV_s$ . 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 $\nu_0 = \frac{3}{2} Q$ .

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