Question

A proton and alpha particle are accelerated through the same potential difference. The ratio of the de-Broglie wavelength of proton to that of alpha particle will be (mass of alpha particle is four times mass of proton.)

• A. $1 : 2$
• B. $2\sqrt{2} : 1$
• C. $1 : 1$
• D. $2 : 1$

Hint:

Remember that an alpha particle is a helium nucleus ($\text{He}^{2+}$), meaning it inherently carries 4 times the mass and 2 times the charge of a standard single proton. Multiplying these factors gives $4 \times 2 = 8$, and taking the square root immediately gives the scaling factor $\sqrt{8} = 2\sqrt{2}$.

Answer 1

The Correct Option is B

Step 1: The comparison.
A proton and an alpha particle are both pushed from rest through the same voltage $V$. We want the ratio of their de-Broglie wavelengths, proton to alpha.
Step 2: The wavelength formula.
A charged particle sped up through voltage $V$ has wavelength \[ \lambda = \frac{h}{\sqrt{2mqV}} \] because the work done $qV$ becomes its kinetic energy.
Step 3: Spot what changes.
Here $h$ and $V$ are the same for both. So the wavelength depends only on mass and charge: \[ \lambda \propto \frac{1}{\sqrt{mq}} \]
Step 4: List masses and charges.
Proton: mass $m$, charge $e$. Alpha: mass $4m$, charge $2e$.
Step 5: Form the ratio.
\[ \frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha q_\alpha}{m_p q_p}} = \sqrt{\frac{(4m)(2e)}{(m)(e)}} = \sqrt{8} \]
Step 6: Simplify.
\[ \sqrt{8} = 2\sqrt{2} \] So the ratio is $2\sqrt{2} : 1$, which is option (2). \[ \boxed{\lambda_p : \lambda_\alpha = 2\sqrt{2} : 1} \]