Question

If the nucleus \(X^{240}\), initially at rest, splits into two daughter nuclei \(Y^{100}\) and \(Z^{140}\), then the ratio of the kinetic energies of \(Y\) and \(Z\) is

• A. \(5:7\)
• B. \(7:5\)
• C. \(1:1\)
• D. \(49:25\)

Hint:

In nuclear fission or decay problems, use conservation of momentum to relate velocities. The kinetic energy ratio of daughter nuclei is inversely proportional to their mass ratio: \[ \frac{K_Y}{K_Z} = \frac{m_Z}{m_Y} \]

Answer 1

The Correct Option is B

Step 1: Understanding the Concept of Conservation of Momentum Since the parent nucleus was initially at rest, the total momentum of the daughter nuclei must be equal and opposite due to conservation of linear momentum: \[ m_Y v_Y = m_Z v_Z \] where: 
- \( m_Y = 100 \) (mass number of nucleus \(Y\)), 
- \( m_Z = 140 \) (mass number of nucleus \(Z\)), 
- \( v_Y \) and \( v_Z \) are their respective velocities. 

Step 2: Kinetic Energy Relation Kinetic energy is given by: \[ K = \frac{1}{2} m v^2 \] Dividing the kinetic energies of the two nuclei: \[ \frac{K_Y}{K_Z} = \frac{\frac{1}{2} m_Y v_Y^2}{\frac{1}{2} m_Z v_Z^2} \] Since \( m_Y v_Y = m_Z v_Z \), we can express velocity as: \[ v_Y = \frac{m_Z}{m_Y} v_Z \] Substituting into the energy equation: \[ \frac{K_Y}{K_Z} = \frac{m_Y \left(\frac{m_Z}{m_Y} v_Z\right)^2}{m_Z v_Z^2} \] \[ = \frac{m_Y m_Z^2 v_Z^2}{m_Y^2 m_Z v_Z^2} \] \[ = \frac{m_Z}{m_Y} = \frac{140}{100} = \frac{7}{5} \] Thus, the correct answer is \( \mathbf{(2)} \ 7:5 \).