Physics · Atomic Physics

JEE Main 2026 — 4 April, Evening Shift — Question 18

The de Broglie wavelength associated with an electron accelerated through a potential difference V is $\lambda_e$ and the de Broglie wavelength associated with a proton accelerated through the same potential difference is $\lambda_p$ . If their corresponding masses are $m_e$ and $m_p$ respectively, then the ratio $\frac{\lambda_e}{\lambda_p}$ is

Option A:
$\sqrt{\frac{m_p}{m_e}}$
Correct
Option B:
$\sqrt{\frac{m_e}{m_p}}$
Option C:
$\frac{m_p}{m_e}$
Option D:
$\left(\frac{m_p}{m_e}\right)^2$

Answer: A

Step-by-step solution

$\lambda = h/\sqrt{2mqV}$ so $\lambda \propto 1/\sqrt{m}$ . Thus $\lambda_e/\lambda_p = \sqrt{m_p/m_e}$ .

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Exam: JEE Main 2026
Paper: 4 April, Evening Shift
Subject: Physics
Chapter: Atomic Physics
Topic: Dual Nature of Matter

In Rutherford's alpha-particle scattering experiment, only a few alpha particles rebound back because: (A) The size of gold nucleus is very…

Answer: A

Step-by-step solution

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