MHTCET 17 Apr 2026 Shift 2 Paper

Concept:The de Broglie wavelength depends on momentum, and the momentum of an accelerated charged particle depends on its mass, charge, and accelerating potential.Explanation:For a charged particle accelerated through a potential difference $V$, the kinetic energy is given by:$$K = qV$$Also, kinetic energy in terms of momentum is:$$K = \frac{p^2}{2m}$$So, the momentum becomes:$$p = \sqrt{2mqV}$$The de Broglie wavelength is:$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}}$$For the $\alpha$-particle and proton to have the same $\lambda$, the product $m q V$ must be equal for both.Thus:$$m_{\alpha} q_{\alpha} V_{\alpha} = m_{p} q_{p} V_{p}$$Here, $m_{\alpha} = 6.4 \times 10^{-27} \, \mathrm{kg}$, $m_p = 1.6 \times 10^{-27} \, \mathrm{kg}$.Charge of $\alpha$-particle: $q_{\alpha} = 2e$; charge of proton: $q_p = e$.Therefore:$$\frac{V_{\alpha}}{V_p} = \frac{m_p q_p}{m_{\alpha} q_{\alpha}}$$$$\frac{V_{\alpha}}{V_p} = \frac{1.6 \times 10^{-27} \times e}{6.4 \times 10^{-27} \times 2e} = \frac{1.6}{12.8} = \frac{1}{8}$$Hence, the ratio is:$$V_{\alpha} : V_p = 1 : 8$$Answer:The correct option is A: $1 : 8$.