Question

The mass of a particle A is double that of particle B and the kinetic energy of B is $ \frac{1}{8} $ that of A. Then the ratio of the de-Broglie wavelength of A to that of B is:

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

Hint:

The de Broglie wavelength is inversely proportional to the momentum. The momentum is related to the square root of the product of mass and kinetic energy.

Answer 1

The Correct Option is C

We are given the following: - The mass of particle A is double that of particle B, so \( m_A = 2 m_B \). 
- The kinetic energy of particle B is \( \frac{1}{8} \) that of A, so \( K_B = \frac{1}{8} K_A \). 
The de Broglie wavelength \( \lambda \) of a particle is given by the formula: \[ \lambda = \frac{h}{p}, \] where \( h \) is Planck's constant and \( p \) is the momentum of the particle. The momentum \( p \) of a particle is related to its kinetic energy \( K \) and mass \( m \) by the equation: \[ K = \frac{p^2}{2m}. \] Thus, the momentum \( p \) can be written as: \[ p = \sqrt{2mK}. \] 
Step 1: Finding the momentum of particles A and B
For particle A, the momentum is: \[ p_A = \sqrt{2m_A K_A} = \sqrt{2(2m_B) K_A} = \sqrt{4m_B K_A} = 2 \sqrt{m_B K_A}. \] For particle B, the momentum is: \[ p_B = \sqrt{2m_B K_B} = \sqrt{2m_B \left( \frac{1}{8} K_A \right)} = \sqrt{\frac{1}{4} m_B K_A} = \frac{1}{2} \sqrt{m_B K_A}. \] 
Step 2: Finding the ratio of the de Broglie wavelengths
Now, the de Broglie wavelength ratio for particles A and B is: \[ \frac{\lambda_A}{\lambda_B} = \frac{p_B}{p_A} = \frac{\frac{1}{2} \sqrt{m_B K_A}}{2 \sqrt{m_B K_A}} = \frac{1}{4}. \] Thus, the ratio of the de Broglie wavelength of A to B is \( 1 : 4 \), which corresponds to option (C).