Question

A proton and an alpha particle are moving with kinetic energies of 4.5 MeV and 0.5 MeV respectively. The ratio of the de Broglie wavelengths of the proton and alpha particle is: 

• A. 2 : 3
• B. 1 : 9
• C. 1 : 2
• D. 1 : 3

Hint:

When comparing de Broglie wavelengths, remember that a heavier particle with the same kinetic energy will have a shorter wavelength. In this case, the alpha particle, being heavier, has a shorter wavelength than the proton when their kinetic energies are considered.

Answer 1

The Correct Option is A

The de Broglie wavelength \(\lambda\) of a particle is given by the relation \(\lambda = \frac{h}{p}\), where \(h\) is the Planck constant and \(p\) is the momentum of the particle. The momentum \(p\) can be related to the kinetic energy \(K\) by the relation \(p = \sqrt{2mK}\), where \(m\) is the mass of the particle.

For the proton:
\[ p_{{proton}} = \sqrt{2m_{{proton}}K_{{proton}}} \]
\[ \lambda_{{proton}} = \frac{h}{\sqrt{2m_{{proton}} \times 4.5 \, {MeV}}} \]

For the alpha particle:
\[ p_{{alpha}} = \sqrt{2m_{{alpha}}K_{{alpha}}} \]
\[ \lambda_{{alpha}} = \frac{h}{\sqrt{2m_{{alpha}} \times 0.5 \, {MeV}}} \]

Since the mass of an alpha particle is approximately four times that of a proton, and considering the kinetic energies, the ratio of their wavelengths becomes:
\[ \frac{\lambda_{{proton}}}{\lambda_{{alpha}}} = \frac{\sqrt{2m_{{alpha}} \times 0.5 \, {MeV}}}{\sqrt{2m_{{proton}} \times 4.5 \, {MeV}}} = \frac{\sqrt{2 \times 4m_{{proton}} \times 0.5}}{\sqrt{2m_{{proton}} \times 4.5}} = \frac{2}{\sqrt{9}} = \frac{2}{3} \]