Question

A particle at rest decays into two particles of mass \( m_1 \) and \( m_2 \) and move with velocities \( v_1 \) and \( v_2 \). The ratio of their de Broglie wave lengths \( \lambda_1 : \lambda_2 \) is:

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

Hint:

The de Broglie wavelength of a particle is inversely proportional to its momentum. For a particle decaying into two smaller particles, their momenta are conserved, so their de Broglie wavelengths are inversely proportional to their masses and velocities.

Answer 1

The Correct Option is B

According to de Broglie's hypothesis, the de Broglie wavelength \( \lambda \) of a particle is given by the equation: \[ \lambda = \frac{h}{p} \] where \( h \) is Planck's constant and \( p \) is the momentum of the particle. For a particle with mass \( m \) and velocity \( v \), the momentum \( p \) is given by: \[ p = mv \] Thus, the de Broglie wavelength for each particle is: \[ \lambda_1 = \frac{h}{m_1 v_1} \quad \text{and} \quad \lambda_2 = \frac{h}{m_2 v_2} \] The ratio of the wavelengths is: \[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{h}{m_1 v_1}}{\frac{h}{m_2 v_2}} = \frac{m_2 v_2}{m_1 v_1} \] Since the initial particle is at rest, by conservation of momentum, the momentum of the two resulting particles is equal and opposite. Therefore, the momentum of each particle must satisfy: \[ m_1 v_1 = m_2 v_2 \] This implies: \[ \frac{m_2 v_2}{m_1 v_1} = 1 \] Thus, the ratio of the de Broglie wavelengths is: \[ \frac{\lambda_1}{\lambda_2} = 1 : 1 \]