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, if they are moving with the same kinetic energy.

Hint:

When dealing with de Broglie wavelength ratios where Kinetic Energy is constant, the ratio simplifies cleanly to the inverse square root of their masses: $\lambda \propto 1/\sqrt{m}$.

Answer 1

Concept:
• The de Broglie wavelength $\lambda$ of a moving particle is intrinsically linked to its momentum $p$ by the equation $\lambda = \frac{h}{p}$, where $h$ is Planck's constant.
• The momentum of a particle can be expressed strictly in terms of its kinetic energy $K$ and mass $m$ using the classical relation $p = \sqrt{2mK}$.
• Combining these yields the functional wavelength formula: $\lambda = \frac{h}{\sqrt{2mK}}$.

Step 1: Establish the formulas for both particles
For the alpha particle (subscript $\alpha$), the wavelength is:
\[ \lambda_\alpha = \frac{h}{\sqrt{2m_\alpha K_\alpha}} \]
For the proton (subscript $p$), the wavelength is:
\[ \lambda_p = \frac{h}{\sqrt{2m_p K_p}} \]

Step 2: Identify the given conditions and mass relationships
The problem explicitly states that both particles possess the exact same kinetic energy, so $K_\alpha = K_p = K$.
We also know the fundamental mass relationship between an alpha particle (a helium nucleus) and a proton: the mass of an alpha particle is approximately four times the mass of a proton.
\[ m_\alpha = 4m_p \]

Step 3: Calculate the required ratio
Divide the alpha particle equation by the proton equation to find the ratio:
\[ \frac{\lambda_\alpha}{\lambda_p} = \frac{\frac{h}{\sqrt{2m_\alpha K}}}{\frac{h}{\sqrt{2m_p K}}} \]
The Planck's constant $h$, the factor of $2$, and the identical kinetic energy $K$ completely cancel out:
\[ \frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{m_p}{m_\alpha}} \]
Substitute the mass relationship $m_\alpha = 4m_p$ into the root:
\[ \frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{m_p}{4m_p}} = \sqrt{\frac{1}{4}} \]
\[ \frac{\lambda_\alpha}{\lambda_p} = \frac{1}{2} \]

Step 4: Conclusion
The specific ratio of their de Broglie wavelengths is strictly 1:2.