Question

A neutron with 0.6MeV kinetic energy directly collides with a stationary carbon nucleus (mass number 12). The kinetic energy of carbon nucleus after the collision is

• A. 1.7 MeV
• B. 0.17 MeV
• C. 17 MeV
• D. Zero

Answer 1

Answer: (B)

0.17 MeV

Answer 2

Kinetic energy transferred to stationary target (carbon

nucleus) $\frac{\Delta K}{K} = \left\lbrack 1 - \left( \frac{m_{1} - m_{2}}{m_{1} + m_{2}} \right)^{2} \right\rbrack$

$\frac{\Delta K}{K} = \left\lbrack 1 - \left( \frac{1 - 12}{1 + 12} \right)^{2} \right\rbrack = \left\lbrack 1 - \frac{121}{169} \right\rbrack = \frac{48}{169}$

∴ $\Delta K = \frac{48}{169} \times (0.6MeV) = 0.17MeV.$