An electron accelerated through a potential difference V₁ has a de-Broglie wavelength of λ. When the potential is changed to V₂, its de-Broglie wavelength increases to 2λ. The value of (V₁/V₂) is equal to (a) 3: 1 (b) 9: 4 (c) 3: 2 (d) 4: 1

Answer: (d). λ = (h)/(√(2meV)) ⇒ λ ∝ 1/√(V), i.e. V ∝ (1)/(λ²). (V₁)/(V₂) = ((λ₂)/(λ₁))² = ((2λ)/(λ))² = 4 So the answer is (d) 4: 1.

An electron accelerated through a potential difference V₁ has a… | MHT CET 2026 (13 Apr, Shift 2)

Q8MHT CET 2026 · 13 APR · SHIFT 2Medium

Matter Waves: de Broglie Wavelength

An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength of $\lambda$ . When the potential is changed to $V_2$ , its de-Broglie wavelength increases to $2\lambda$ . The value of $(V_1/V_2)$ is equal to

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Answer: (d)

$\lambda = \dfrac{h}{\sqrt{2meV}} \Rightarrow \lambda \propto \dfrac{1}{\sqrt{V}}$ , i.e. $V \propto \dfrac{1}{\lambda^2}$ .
$\dfrac{V_1}{V_2} = \left(\dfrac{\lambda_2}{\lambda_1}\right)^2 = \left(\dfrac{2\lambda}{\lambda}\right)^2 = 4$
So the answer is (d) $4 : 1$ .

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