Question

The ratio of de-Broglie wavelength of an $\alpha$-particle and proton accelerated from rest by the same potential is $\dfrac{1}{\sqrt{m}}$. The value of $m$ is

• A. $2$
• B. $8$
• C. $4$
• D. $5$

Hint:

De Broglie wavelength after acceleration through V is h over root of 2 m q V.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
A particle of mass \(m\) and charge \(q\) accelerated from rest through a potential \(V\) has momentum \(p=\sqrt{2mqV}\). So \(\lambda=\dfrac{h}{\sqrt{2mqV}}\).

Step 2: Ratio
Same \(V\) for both:
\[ \frac{\lambda_\alpha}{\lambda_p}=\sqrt{\frac{m_pq_p}{m_\alpha q_\alpha}} \]

Step 3: Insert values
An alpha particle has mass about 4 times that of a proton and charge 2 times:
\[ \frac{\lambda_\alpha}{\lambda_p}=\sqrt{\frac{1\times1}{4\times2}}=\frac1{\sqrt8} \]

Step 4: Compare
The given ratio is \(\dfrac1{\sqrt m}\), so \(m=8\), option (B).

Final Answer:
The alpha has 4 times the mass and twice the charge, so the wavelength ratio is 1 over root 8 and m = 8, option (B). \[ \boxed{8} \]