For a certain metal, when monochromatic light of wavelength λ is incident, the stopping potential for photoelectrons is 3V₀. When the same metal is illuminated by light of wavelength 2λ, then the stopping potential becomes V₀. The threshold wavelength for photoelectric emission for the given metal is αλ. The value of α is ______. (a) 1 (b) 4 (c) 2 (d) 3

Answer: (b). Einstein: (hc)/(λ) = φ + eVs; write A = (hc)/(λ). A - φ = 3eV₀ and A/2 - φ = eV₀ Subtract: A/2 = 2eV₀ ⇒ A = 4eV₀, φ = eV₀ = A/4 λ₀ = (hc)/(φ) = 4λ So the answer is (b) α = 4.

For a certain metal, when monochromatic light of wavelength λ is… | JEE Main 2026 (2 Apr, Shift 1)

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Photoelectric Effect: Einstein's Equation

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 ______.

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Answer: (b)

Einstein: $\frac{hc}{\lambda} = \phi + eV_s$ ; write $A = \frac{hc}{\lambda}$ .
$A - \phi = 3eV_0$ and $\frac{A}{2} - \phi = eV_0$
Subtract: $\frac{A}{2} = 2eV_0 \Rightarrow A = 4eV_0,\ \phi = eV_0 = \frac{A}{4}$
$\lambda_0 = \frac{hc}{\phi} = 4\lambda$
So the answer is (b) $\alpha = 4$ .

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