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 by 50%. The value of ((V₁)/(V₂)) is equal to (a) 3/2 (b) 4 (c) 3 (d) 9/4

Answer: (d). λ ∝ 1/√(V) ⇒ (V₁)/(V₂) = ((λ₂)/(λ₁))². λ₂ = 1.5λ ⇒ (V₁)/(V₂) = (1.5)² = 9/4 So the answer is (d) 9/4.

An electron accelerated through a potential difference V₁ has a… | JEE Main 2023 (30 Jan, Shift 2)

Q74JEE Main 2023 · 30 JAN · 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 by $50\%$ . The value of $\left(\frac{V_1}{V_2}\right)$ is equal to

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Answer: (d)

$\lambda \propto \dfrac{1}{\sqrt{V}} \Rightarrow \dfrac{V_1}{V_2} = \left(\dfrac{\lambda_2}{\lambda_1}\right)^2$ .
$\lambda_2 = 1.5\lambda \Rightarrow \dfrac{V_1}{V_2} = (1.5)^2 = \dfrac{9}{4}$
So the answer is (d) $\dfrac{9}{4}$ .

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