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: Approach
Use kinetic energy instead of momentum.

Step 2: Kinetic energy
Both gain $K=qV$, so the alpha gets twice the proton's energy. Using $\lambda=\dfrac{h}{\sqrt{2mK}}$, the product $mK$ is $4\times2=8$ times larger for the alpha.

Step 3: Ratio
$\lambda$ is smaller by $\sqrt8$, so $\dfrac{\lambda_\alpha}{\lambda_p}=\dfrac{1}{\sqrt8}$ and $m=8$. 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). \[ \boxed{8} \]