QJEEM23Ph75aa: JEE Main 2023 Physics - Class 12 JEEM23Ph75aa Solution: A proton and an -particle are accelerated from rest by and…

The de-Broglie wavelength of a particle is given by:

$\lambda = \frac{h}{p}$

where $h$ is the Planck's constant and $p$ is the momentum of the particle.

The momentum of a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by:

$p = \sqrt{2 m q V}$

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 $q = 2 \times 1.6 \times 10^{-19} \mathrm{~C}$ .

Therefore, the ratio of their de-Broglie wavelengths is:

$\frac{\lambda_p}{\lambda_\alpha} = \frac{p_\alpha}{p_p} = \sqrt{\frac{m_\alpha}{m_p} \cdot \frac{q_\alpha V_\alpha}{q_p V_p}} = \sqrt{\frac{6.64 \times 10^{-27}}{1.67 \times 10^{-27}} \cdot \frac{2 \times 1.6 \times 10^{-19} \times 4}{1.6 \times 10^{-19} \times 2}} = \sqrt{16} = 4$

Therefore, the ratio of their de-Broglie wavelengths is 4 : 1.