Question

A proton and an alpha particle have equal momentum. The ratio of their kinetic energies (\(E_p /E_\alpha\)) and de Broglie wavelengths associated with them (\(\lambda_p /\lambda_\alpha\)) respectively are :

• A. 2, 1
• B. 1, 2
• C. 4, 1
• D. 1, 4

Hint:

Since the de Broglie wavelength depends solely on momentum (\(\lambda = \frac{h}{p}\)), any two particles with equal momentum will always have identical de Broglie wavelengths, meaning their wavelength ratio must be 1. This helps you eliminate options immediately.

Answer 1

The Correct Option is C

Concept: The motion of a particle can be characterized using both its momentum (\(p\)) and its kinetic energy (\(E\)). The fundamental algebraic relationship connecting kinetic energy and linear momentum is: \[ E = \frac{p^2}{2m} \] Additionally, according to the de Broglie hypothesis, every moving particle exhibits wave-like characteristics, and its associated de Broglie wavelength (\(\lambda\)) is given by: \[ \lambda = \frac{h}{p} \] where \(h\) is Planck's constant.

Step 1: Identifying properties of a proton and an alpha particle.
Let us establish the mass metrics of both subatomic particles relative to a basic atomic mass unit:
• For a proton (\(p\)): Mass is \(m_p = m\).
• For an alpha particle (\(\alpha\), which is a helium nucleus containing 2 protons and 2 neutrons): Mass is \(m_\alpha = 4m\). The problem states that both particles have equal linear momentum: \[ p_p = p_\alpha = p \]

Step 2: Evaluating the ratio of their kinetic energies (\(E_p / E_\alpha\)).
Using the kinetic energy-momentum relation: \[ E_p = \frac{p^2}{2m_p} = \frac{p^2}{2m} \] \[ E_\alpha = \frac{p^2}{2m_\alpha} = \frac{p^2}{2(4m)} = \frac{p^2}{8m} \] Taking the ratio of \(E_p\) to \(E_\alpha\): \[ \frac{E_p}{E_\alpha} = \frac{\frac{p^2}{2m}}{\frac{p^2}{8m}} = \frac{8m}{2m} = 4 \] Thus, the kinetic energy ratio is 4.

Step 3: Evaluating the ratio of their de Broglie wavelengths (\(\lambda_p / \lambda_\alpha\)).
Using the de Broglie wavelength relation: \[ \lambda_p = \frac{h}{p_p} = \frac{h}{p} \] \[ \lambda_\alpha = \frac{h}{p_\alpha} = \frac{h}{p} \] Taking the ratio of \(\lambda_p\) to \(\lambda_\alpha\): \[ \frac{\lambda_p}{\lambda_\alpha} = \frac{\frac{h}{p}}{\frac{h}{p}} = 1 \] Therefore, the kinetic energy ratio is 4 and the wavelength ratio is 1, which corresponds to Option (C).