The de Broglie wavelength for an electron accelerated through the potential difference V 1 volt is λ 1 . When the potential difference is changed to V 2 volt, the associated de Broglie wavelength is increased by 50\% . If (V 1/V 2) = (9/α) , then the value of α is .

Physics · Atomic Physics

JEE Main 2026 — 6 April, Evening Shift — Question 17

The de Broglie wavelength for an electron accelerated through the potential difference $V_1$ volt is $\lambda_1$ . When the potential difference is changed to $V_2$ volt, the associated de Broglie wavelength is increased by $50\%$ . If $(V_1/V_2) = (9/\alpha)$ , then the value of $\alpha$ is ______.

Answer: 4

Numerical answer — enter this value.

Step-by-step solution

$\lambda \propto 1/\sqrt{V}$ . Given $\lambda_2 = 1.5\lambda_1$ , so $\sqrt{V_1/V_2} = 1.5 \Rightarrow V_1/V_2 = 2.25 = 9/4$ . Thus $\alpha = 4$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 6 April, Evening Shift
Subject: Physics
Chapter: Atomic Physics
Topic: Dual Nature of Matter