An electron accelerated through a potential difference ' V₁ ' has a de-Broglie wavelength ' λ '. When the potential is changed to ' V₂ ' its de-Broglie wavelength increases by 50%. The value of ((V₁)/(V₂)) is (a) 3: 1 (b) 9: 4 (c) 3: 2 (d) 4: 1

Answer: (b). λ ∝ 1/√(V) ⇒ (V₁)/(V₂) = ((λ₂)/(λ₁))². A 50% increase gives λ₂ = 1.5λ. (V₁)/(V₂) = (1.5)² = 9/4 So the answer is (b) 9: 4.

An electron accelerated through a potential difference ' V₁ ' has a… | MHT CET 2023 (10 May, Shift 1)

Q73MHT CET 2023 · 10 MAY · SHIFT 1Medium

Matter Waves: de Broglie Wavelength

An electron accelerated through a potential difference ' $V_1$ ' has a de-Broglie wavelength ' $\lambda$ '. 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

Show answer and solution

Answer: (b)

$\lambda \propto \dfrac{1}{\sqrt{V}} \Rightarrow \dfrac{V_1}{V_2} = \left(\dfrac{\lambda_2}{\lambda_1}\right)^2$ .
A $50\%$ increase gives $\lambda_2 = 1.5\lambda$ .
$\dfrac{V_1}{V_2} = (1.5)^2 = \dfrac{9}{4}$
So the answer is (b) $9 : 4$ .

Report a mistake

Matter Waves: de Broglie Wavelength question from MHT CET 2023 (10 May, Shift 1)

More questions on Matter Waves: de Broglie Wavelength