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

The correct option is 0.17 MeV.

To find the kinetic energy of the carbon nucleus after a direct (head-on), elastic collision with a neutron, we can use the principles of conservation of momentum and conservation of kinetic energy.

Let:
- Mass of the neutron, $m=1\text{u}$
- Mass of the carbon nucleus, $M=12\text{u}$
- Initial kinetic energy of the neutron, $K_i=0.6\text{MeV}$
- Initial velocity of the carbon nucleus, $V=0$ (since it is stationary)

For a perfectly elastic head-on collision between a moving mass $m$ and a stationary target mass $M$, the kinetic energy transferred to the target mass ($K_{\text{target}}$) is given by the formula:

$K_{\text{target}}=\left[\frac{4mM}{{(m+M)}^2}\right]\times K_i$

Substituting the given values into the formula:

$K_{\text{carbon}}=\frac{4\times 1\times 12}{{(1+12)}^2}\times 0.6\text{MeV}$

Simplify the expression step-by-step:

$K_{\text{carbon}}=\frac{48}{{13}^2}\times 0.6\text{MeV}$

$K_{\text{carbon}}=\frac{48}{169}\times 0.6\text{MeV}$

$K_{\text{carbon}}\approx 0.284\times 0.6\text{MeV}$

$K_{\text{carbon}}\approx 0.1704\text{MeV}$

Rounding to two decimal places, the kinetic energy of the carbon nucleus after the collision is 0.17 MeV.