Question

The proton and \(α\) - particle are accelerated through same potential difference. Then the ratio of the de-Broglie wavelength of proton and \(α\) - particle is (mass of \(α\)-particle is \(4\) times mass of proton, charge of \(α\)-particle is \(2\) times charge of proton)

• A. \(3\sqrt{3}\)
• B. \(3\sqrt{2}\)
• C. \(2\sqrt{3}\)
• D. \(2\sqrt{2}\)

Hint:

lambda = h / sqrt(2 m q V) for a particle accelerated through V.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A charge \(q\) accelerated through potential difference \(V\) gains kinetic energy \(qV\), so its momentum is \(p = \sqrt{2mqV}\). The de Broglie wavelength is
\[ \lambda = \frac hp = \frac{h}{\sqrt{2mqV}} \]

Step 2: Ratio:
Same \(V\) for both, so \(\lambda\propto\frac{1}{\sqrt{mq}}\):
\[ \frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha q_\alpha}{m_pq_p}} = \sqrt{4\times2} = \sqrt8 = 2\sqrt2 \]

Step 3: Why the other options are wrong.
\(3\sqrt3\) and \(3\sqrt2\) involve a factor of 3, which does not appear in the mass or charge ratios. \(2\sqrt3\) would need \(mq\) ratio 12.

Final Answer:
The ratio is \(2\sqrt2\), option (D). \[ \boxed{2\sqrt{2}} \]