JEE Main 2026 · 2 Apr · Shift 1 · Physics Q42

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PhysicsSingle correct+4 / -1Question 42 of 75

For a certain metal, when monochromatic light of wavelength $\lambda$ is incident, the stopping potential for photoelectrons is $3V_0$ . When the same metal is illuminated by light of wavelength $2\lambda$ , then the stopping potential becomes $V_0$ . The threshold wavelength for photoelectric emission for the given metal is $\alpha\lambda$ . The value of $\alpha$ is ________.

Answer and worked solution

Official answer: B

Einstein equation for both wavelengths.
$\frac{hc}\lambda-\phi=3eV_0,\qquad\frac{hc}{2\lambda}-\phi=eV_0.$

Subtract. $\frac{hc}{2\lambda}=2eV_0$ , so $eV_0=\frac{hc}{4\lambda}$ and $\phi=\frac{hc}{2\lambda}-\frac{hc}{4\lambda}=\frac{hc}{4\lambda}$ .

Threshold. $\lambda_0=\dfrac{hc}\phi=4\lambda$ , so $\alpha=4$ .

(B)

Topics:Photoelectric effect

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JEE Main 2026 · 2 Apr · Shift 1, Physics question 42