Question

An $\alpha$ particle and a proton are accelerated from rest by a potential difference of $200V$. After this, their de Broglie wavelengths are ${\lambda }_{\alpha }$ and ${\lambda }_p$ respectively. The ratio $\frac{{\lambda }_p}{{\lambda }_{\alpha }}$ is:

• A. $8$
• B. $2.8$
• C. $3.8$
• D. $7.8$

Answer 1

The correct option is B

$2.8$

Explanation for Correct answer:

B) $\mathbf{2}\mathbf{.}\mathbf{8}$

1. The potential difference between accelerated $\mathrm{\alpha }$- particle and proton is $200V$ and the De Broglie wavelength of the particles is ${\mathrm{\lambda }}_{\mathrm{\alpha }}$ and ${\mathrm{\lambda }}_{\mathrm{p}}$.
2. We know that $KE$ $=qv$ where $\mathrm{q}$ is charge and $\mathrm{v}$ is potential difference.
3. Now, De Broglie wavelength $=\frac{h}{p}=\frac{h}{\sqrt{2mKE}}=\frac{h}{\sqrt{2mqv}}$
4. Therefore the ratio of De Broglie wavelength $=\frac{{\lambda }_p}{{\lambda }_{\alpha }}=\frac{\sqrt{m_{\alpha }{q}_{\alpha }}}{\sqrt{m_p{q}_p}}$
5. $\alpha$-particle has four times the mass and twice the charge of a proton; $=\frac{{\lambda }_p}{{\lambda }_{\alpha }}=\frac{\sqrt{4m_{p}2q_p}}{\sqrt{m_p{q}_p}}$
6. So, $\frac{{\lambda }_p}{{\lambda }_{\alpha }}=\sqrt{8}=2.8$

Explanation for incorrect answers:

A) $\mathbf{8}$

• The ratio of De Broglie wavelength, $\frac{{\lambda }_p}{{\lambda }_{\alpha }}=2.8$

C) $\mathbf{3}\mathbf{.}\mathbf{8}$

• The ratio of De Broglie wavelength, $\frac{{\lambda }_p}{{\lambda }_{\alpha }}=2.8$

D) $\mathbf{7}\mathbf{.}\mathbf{8}$

• The ratio of De Broglie wavelength, $\frac{{\lambda }_p}{{\lambda }_{\alpha }}=2.8$

Hence, option B is correct, the ratio of De Broglie wavelength, $\frac{{\lambda }_p}{{\lambda }_{\alpha }}=2.8$