JEE Main 2026 — Apr 4, Shift 1 — Physics solution

JEE Main2026Apr 4, Shift 1Physics

The graph shows variation of stopping potential $V_{0}$ with the frequency $ν$ of the incident radiation for three photosensitive metals $X_{1}$ , $X_{2}$ and $X_{3}$ . Which metal will give out electrons with greater kinetic energy, for the same wavelength of incident radiation?

A
$X_{1}$
B
$X_{2}$
C
$X_{3}$
D
All the metals will give out photo electrons with same kinetic energies.

Solution

From Einstein's photoelectric equation, $K_{m a x} = h ν - ϕ$ . For a fixed incident frequency (fixed wavelength), the metal with the smallest work function $ϕ$ gives the largest $K_{m a x}$ .

From the graph, the $x$ -intercept of each line equals the threshold frequency $ν_{0} = ϕ / h$ . The threshold frequencies are in the ratio $X_{1} : X_{2} : X_{3} = 1.0 : 1.5 : 2.0$ , so $X_{1}$ has the smallest $ϕ$ .

Therefore $X_{1}$ emits photoelectrons with the greatest kinetic energy.

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