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: Product of mass and charge:
Alpha particle: $m_\alpha q_\alpha = (4m)(2e) = 8me$. Proton: $me$.

Step 2: Wavelength inversely depends on the root of this product:
The alpha particle has $\sqrt8$ times the root, so its wavelength is smaller by that factor.

Step 3: Ratio:
$\lambda_p/\lambda_\alpha = \sqrt8 = 2\sqrt2$.

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