Question

Draw a graph between de Broglie wavelength (\(\lambda\)) and momentum (\(p\)) of a moving particle. A proton and an \(\alpha\)-particle are accelerated by the same potential. Compute the ratio of their de Broglie wavelengths.

Hint:

The de Broglie wavelength of a particle is inversely proportional to its momentum. The ratio of de Broglie wavelengths for two particles can be calculated by finding the ratio of their momenta, considering their masses and charges.

Answer 1

The de Broglie wavelength (\(\lambda\)) of a particle is related to its momentum (\(p\)) by the equation:
\[ \lambda = \frac{h}{p} \] Where:
- \(\lambda\) is the de Broglie wavelength,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J} . \text{s}\)),
- \(p\) is the momentum of the particle.
Since momentum is given by \(p = mv\), where \(m\) is the mass and \(v\) is the velocity of the particle, we can write:
\[ \lambda = \frac{h}{mv} \] Now, consider a proton and an \(\alpha\)-particle (helium nucleus) accelerated by the same potential difference. The kinetic energy \(K\) acquired by a particle when accelerated by a potential \(V\) is given by:
\[ K = \frac{1}{2} mv^2 = qV \] Where \(q\) is the charge of the particle and \(V\) is the potential. For a proton, \(q = e\) and for an \(\alpha\)-particle, \(q = 2e\). From this, we can find the velocity of each particle and subsequently the momentum.
For the proton, the velocity \(v_p\) is:
\[ \frac{1}{2} m_p v_p^2 = eV ⇒ v_p = \sqrt{\frac{2eV}{m_p}} \] For the \(\alpha\)-particle, the velocity \(v_{\alpha}\) is:
\[ \frac{1}{2} m_{\alpha} v_{\alpha}^2 = 2eV ⇒ v_{\alpha} = \sqrt{\frac{4eV}{m_{\alpha}}} \] Now, the de Broglie wavelength for the proton (\(\lambda_p\)) and the \(\alpha\)-particle (\(\lambda_{\alpha}\)) are:
\[ \lambda_p = \frac{h}{m_p v_p} = \frac{h}{m_p \sqrt{\frac{2eV}{m_p}}} = \frac{h}{\sqrt{2 m_p e V}} \] \[ \lambda_{\alpha} = \frac{h}{m_{\alpha} v_{\alpha}} = \frac{h}{m_{\alpha} \sqrt{\frac{4eV}{m_{\alpha}}}} = \frac{h}{\sqrt{4 m_{\alpha} e V}} \] Finally, the ratio of their de Broglie wavelengths is:
\[ \frac{\lambda_p}{\lambda_{\alpha}} = \frac{\frac{h}{\sqrt{2 m_p e V}}}{\frac{h}{\sqrt{4 m_{\alpha} e V}}} = \sqrt{\frac{4 m_{\alpha}}{2 m_p}} = \sqrt{2 \frac{m_{\alpha}}{m_p}} \] Since the mass of a proton \(m_p\) is approximately 1836 times the mass of an electron, and the mass of an \(\alpha\)-particle \(m_{\alpha} = 4 m_p\), we get:
\[ \frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{2 \times \frac{4 m_p}{m_p}} = \sqrt{8} \approx 2.83 \] Thus, the ratio of the de Broglie wavelengths of the proton and \(\alpha\)-particle is approximately 2.83.