Question

A proton and an \( \alpha \)-particle are accelerated by the same potential difference. Find out the ratio of their de Broglie wavelengths.

Hint:

For de Broglie wavelengths, remember that the momentum \( p \) is proportional to \( \sqrt{mV} \), so heavier particles or particles with higher charges will have smaller wavelengths.

Answer 1

The de Broglie wavelength \( \lambda \) of a particle is given by the formula: \[ \lambda = \frac{h}{p} \] Where: - \( h \) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J} . \text{s}\)), - \( p \) is the momentum of the particle.
The momentum \( p \) of a particle accelerated by a potential difference \( V \) is given by: \[ p = \sqrt{2m e V} \] Where: - \( m \) is the mass of the particle,
- \( e \) is the charge of the particle (\( 1.6 \times 10^{-19} \, \text{C} \)),
- \( V \) is the potential difference.
Now, we calculate the momentum for both the proton and the \( \alpha \)-particle.
1. For the proton:
- Mass of proton \( m_p = 1.67 \times 10^{-27} \, \text{kg} \),
- Charge of proton \( e_p = 1.6 \times 10^{-19} \, \text{C} \).
The momentum of the proton is: \[ p_p = \sqrt{2 \times 1.67 \times 10^{-27} \times 1.6 \times 10^{-19} \times V} \] 2. For the \( \alpha \)-particle:
- Mass of \( \alpha \)-particle \( m_{\alpha} = 4 \times 1.67 \times 10^{-27} \, \text{kg} = 6.68 \times 10^{-27} \, \text{kg} \),
- Charge of \( \alpha \)-particle \( e_{\alpha} = 2 \times 1.6 \times 10^{-19} \, \text{C} \).
The momentum of the \( \alpha \)-particle is: \[ p_{\alpha} = \sqrt{2 \times 6.68 \times 10^{-27} \times 2 \times 1.6 \times 10^{-19} \times V} \] 3. Ratio of the de Broglie wavelengths:}
Since the de Broglie wavelength is inversely proportional to the momentum, the ratio of the wavelengths is: \[ \frac{\lambda_p}{\lambda_{\alpha}} = \frac{p_{\alpha}}{p_p} = \frac{\sqrt{2 \times 6.68 \times 10^{-27} \times 2 \times 1.6 \times 10^{-19} \times V}}{\sqrt{2 \times 1.67 \times 10^{-27} \times 1.6 \times 10^{-19} \times V}} \] Simplifying the ratio: \[ \frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{6.68 \times 2}{1.67}} = \sqrt{8} = 2.83 \] Thus, the ratio of the de Broglie wavelengths is approximately 2.83.