Question

Find ratio ($\lambda_a/\lambda_p$) of the de Broglie wavelength $\lambda_a$ and $\lambda_p$ associated respectively with an alpha particle and a proton, just after they are accelerated through the same potential difference.

Hint:

For particles accelerated through the same voltage, the wavelength relies purely on the inverse square root of the product of mass and charge: $\lambda \propto 1/\sqrt{mq}$. Always memorize the alpha particle's properties ($4m_p$, $2q_p$) as they are heavily tested.

Answer 1

Concept:
• When a charged particle of charge $q$ is aggressively accelerated from rest across a specific potential difference $V$, it naturally acquires a kinetic energy mathematically equal to $K = qV$.
• Substituting this energy formulation into the standard de Broglie relation $\lambda = \frac{h}{\sqrt{2mK}}$ yields a new functional equation: $\lambda = \frac{h}{\sqrt{2mqV}}$.

Step 1: Establish the formulas for both particles
For the alpha particle, the wavelength after acceleration is:
\[ \lambda_\alpha = \frac{h}{\sqrt{2m_\alpha q_\alpha V_\alpha}} \]
For the proton, the corresponding wavelength is:
\[ \lambda_p = \frac{h}{\sqrt{2m_p q_p V_p}} \]

Step 2: Identify the given conditions, mass, and charge relationships
The problem explicitly states they are accelerated through the identical potential difference, meaning $V_\alpha = V_p = V$.
The mass of an alpha particle is four times that of a proton: $m_\alpha = 4m_p$.
The physical charge of an alpha particle (two protons) is precisely twice the charge of a single proton: $q_\alpha = 2q_p$.

Step 3: Calculate the required ratio
Divide the alpha particle equation strictly by the proton equation to cleanly find the ratio:
\[ \frac{\lambda_\alpha}{\lambda_p} = \frac{\frac{h}{\sqrt{2m_\alpha q_\alpha V}}}{\frac{h}{\sqrt{2m_p q_p V}}} \]
The physical constants $h$, $2$, and the identical potential difference $V$ completely cancel out:
\[ \frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{m_p q_p}{m_\alpha q_\alpha}} \]
Substitute the known mass and charge relationships into the mathematical root:
\[ \frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{m_p \cdot q_p}{(4m_p) \cdot (2q_p)}} \]
The mass and charge variables cancel out elegantly:
\[ \frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{1}{4 \times 2}} = \sqrt{\frac{1}{8}} \]
\[ \frac{\lambda_\alpha}{\lambda_p} = \frac{1}{2\sqrt{2}} \]

Step 4: Conclusion
The specific ratio of their de Broglie wavelengths after being accelerated through the same potential is exactly $1:2\sqrt{2}$.