… and … be the maximum kinetic energies…

Single correct+4 / -1Answer matches NTA’s final key

$K_1$ and $K_2$ be the maximum kinetic energies of photoelectrons emitted from a surface of a given material for the light of wavelength $\lambda_1$ and $\lambda_2$ , respectively. If $\lambda_1 = 2\lambda_2$ then the work function of material is given by:

$K_2 + 2K_1$
$2K_2 - K_1$
$K_1 - 2K_2$
$K_2 - 2K_1$

Solution

Write $E = \frac{hc}{\lambda_1}$ . Since $\lambda_2 = \frac{\lambda_1}{2}$ , the second photon carries $2E$ :

$K_1 = E - \phi, \qquad K_2 = 2E - \phi.$

Eliminate $E$ : $K_2 - 2K_1 = -\phi + 2\phi = \phi$ .

Answer: $\phi = K_2 - 2K_1$ .

Answer checked against the official answer key. The solution is our own working.

JEE MAIN 2026 · Dual Nature of Matter and Radiation

Solution

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