Question

An $\alpha$ particle and a proton are accelerated from rest by a potential difference of 200 V. After this, their de Broglie wavelengths are $\lambda_\alpha$ and $\lambda_p$ respectively. The ratio $\lambda_p/\lambda_\alpha$ is :

• A. 2.8
• B. 8
• C. 7.8
• D. 3.8

Hint:

When particles are accelerated by the {same potential}, the wavelength ratio depends only on the square root of the product of their mass and charge: $\lambda \propto \frac{1}{\sqrt{mq}}$.

Answer 1

The Correct Option is A

Step 1: The de Broglie wavelength for a particle accelerated by potential $V$ is $\lambda = \frac{h}{\sqrt{2mqV}}$.
Step 2: For the ratio: $\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha q_\alpha}{m_p q_p}}$.
Step 3: Known values: $m_\alpha = 4m_p$ and $q_\alpha = 2q_p$.
Step 4: $\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{4m_p \times 2q_p}{m_p \times q_p}} = \sqrt{8} = 2\sqrt{2}$.
Step 5: $2 \times 1.414 = 2.828 \approx 2.8$.