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

The de-Broglie wavelength of a particle is given by:<br/><br/> $\lambda = \frac{h}{p}$<br/><br/> where $h$ is the Planck's constant and $p$ is the momentum of the particle. <br/><br/> The momentum of a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by:<br/><br/> $p = \sqrt{2 m q V}$ <br/><br/> For a proton, $m = 1.67 \times 10^{-27} \mathrm{~kg}$ and $q = 1.6 \times 10^{-19} \mathrm{~C}$, and for an $\alpha$-particle, $m = 6.64 \times 10^{-27} \mathrm{~kg}$ and $q = 2 \times 1.6 \times 10^{-19} \mathrm{~C}$. <br/><br/> Therefore, the ratio of their de-Broglie wavelengths is:<br/><br/> $$\frac{\lambda_p}{\lambda_\alpha} = \frac{p_\alpha}{p_p} = \sqrt{\frac{m_\alpha}{m_p} \cdot \frac{q_\alpha V_\alpha}{q_p V_p}} = \sqrt{\frac{6.64 \times 10^{-27}}{1.67 \times 10^{-27}} \cdot \frac{2 \times 1.6 \times 10^{-19} \times 4}{1.6 \times 10^{-19} \times 2}} = \sqrt{16} = 4$$ <br/><br/> Therefore, the ratio of their de-Broglie wavelengths is 4 : 1.