Question

A proton (p) and an $\alpha$-particle are accelerated to the same potential difference. Find the ratio of de Broglie wavelengths associated with proton and $\alpha$-particle.

Hint:

Heavier and more charged particles have smaller de Broglie wavelengths.

Answer 1

Step 1: De Broglie relation.
\[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2m q V}} \] where $m$ = mass, $q$ = charge, $V$ = accelerating potential.
Step 2: For proton.
\[ \lambda_p = \frac{h}{\sqrt{2 m_p e V}} \]
Step 3: For $\alpha$-particle.
Mass $m_\alpha = 4m_p$, charge $q_\alpha = 2e$. \[ \lambda_\alpha = \frac{h}{\sqrt{2 (4m_p)(2e)V}} = \frac{h}{\sqrt{16 m_p e V}} \]
Step 4: Ratio.
\[ \frac{\lambda_p}{\lambda_\alpha} = \frac{\dfrac{h}{\sqrt{2 m_p e V}}}{\dfrac{h}{\sqrt{16 m_p e V}}} = \sqrt{\frac{16}{2}} = \sqrt{8} = 2.83 \] If approximated as integer ratio: $\dfrac{\lambda_p}{\lambda_\alpha} \approx 3$.
Step 5: Conclusion.
The ratio $\lambda_p : \lambda_\alpha \approx 2.8 : 1 \, \approx 3 : 1$.