MHTCET 20 Apr 2026 Shift 1 Paper

Concept:For a charged particle accelerated through a potential difference $V$, its de-Broglie wavelength depends on mass and charge as $\lambda = \frac{h}{\sqrt{2mqV}}$.Explanation:The kinetic energy gained by a particle of charge $q$ accelerated through potential $V$ is:$$qV = \frac{p^2}{2m}$$So the momentum is:$$p = \sqrt{2mqV}$$Hence, the de-Broglie wavelength is:$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}}$$Since both particles are accelerated through the same potential difference $V$, $h$ and $V$ cancel when taking the ratio:$$\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha q_\alpha}{m_p q_p}}$$Given $m_\alpha = 4m_p$ and $q_\alpha = 2q_p$, we get:$$\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{4m_p \times 2q_p}{m_p q_p}} = \sqrt{8} = 2\sqrt{2}$$Answer:The ratio of the de-Broglie wavelength of proton to $\alpha$-particle is $2\sqrt{2}$, which is Option D.