A proton and an -particle are accelerated from rest by 2V and 4V… — Solution | JEE Physics

MCQEasyJEE 2023de Broglie Relation

A proton and an $\alpha$ -particle are accelerated from rest by $2V$ and $4V$ potentials, respectively. The ratio of their de-Broglie wavelength is:

Answer

Correct answer:A

Standard Method

Given: $V_p = 2V$ , $V_{\alpha} = 4V$ , $q_p = e$ , $q_{\alpha} = 2e$ , $m_p = m_p$ , $m_{\alpha} = 4m_p$ .

Find: The ratio $\frac{\lambda_p}{\lambda_{\alpha}}$ .

For a particle accelerated through potential $V$ , de-Broglie wavelength varies as

\lambda = \frac{h}{\sqrt{2mqV}}

Therefore,

\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha} q_{\alpha} V_{\alpha}}{m_p q_p V_p}}

Substituting the given values,

\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{4m_p \cdot 2e \cdot 4V}{m_p \cdot e \cdot 2V}}

\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{32m_p e V}{2m_p e V}}

\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{16} = 4

Therefore, the ratio of their de-Broglie wavelengths is $4:1$ . The correct option is A.

Using $\lambda \propto \sqrt{qV/m}$ instead of the correct relation $\lambda \propto 1/\sqrt{mqV}$ . This reverses the ratio. Start from $\lambda = h/\sqrt{2mqV}$ before substituting.
Taking the $\alpha$ -particle charge as $e$ instead of $2e$ . This is wrong because an $\alpha$ -particle has two protons. Use $q_{\alpha} = 2e$ .
Taking the mass of the $\alpha$ -particle as $2m_p$ instead of $4m_p$ . This gives an incorrect ratio. Use the standard approximation $m_{\alpha} = 4m_p$ .

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.