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: Understanding the Question:
The problem requires finding the ratio of de-Broglie wavelengths ($\lambda_p : \lambda_\alpha$) for a proton and an alpha particle. Both particles start from rest and are accelerated through an identical electrical potential difference ($V$).

Step 2: Key Formula or Approach:
The de-Broglie wavelength $\lambda$ of a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by: $$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE_k}} = \frac{h}{\sqrt{2mqV}}$$ Since Planck's constant $h$ and the accelerating potential $V$ are constant for both particles, the wavelength is inversely proportional to the square root of the product of mass and charge: $$\lambda \propto \frac{1}{\sqrt{mq}} \implies \frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha q_\alpha}{m_p q_p}}$$

Step 3: Detailed Explanation:
Establish the relative physical properties of a proton ($p$) and an alpha particle ($\alpha$): Mass relations: $m_p = m$ and $m_\alpha = 4m$ Charge relations: $q_p = e$ and $q_\alpha = 2e$
Substitute these mass and charge variables into our inverse ratio formula: $$\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{(4m)(2e)}{(m)(e)}} = \sqrt{\frac{8me}{me}}$$ Cancel out the common variable parameters $m$ and $e$ from the radical expression: $$\frac{\lambda_p}{\lambda_\alpha} = \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$$ Expressing this as a standard notation ratio yields $2\sqrt{2} : 1$.

Step 4: Final Answer:
The ratio of the de-Broglie wavelengths is $2\sqrt{2} : 1$, which corresponds precisely to option (B).