A neutron with a kinetic energy of 12.5 eV strikes a stationary...

Let the mass of the neutron and the hydrogen atom be $m$.

Initial kinetic energy of the neutron, $K = 12.5 \text{ eV}$.

The maximum energy that can be lost in the collision occurs when the collision is perfectly inelastic.

Maximum loss in kinetic energy, \Delta K_\max = \dfrac{1}{2} \mu v_{rel}^2 = \dfrac{1}{2} \left( \dfrac{m \cdot m}{m + m} \right) v^2 = \dfrac{1}{2} \left( \dfrac{1}{2} m v^2 \right) = \dfrac{K}{2}

\Delta K_\max = \dfrac{12.5}{2} = 6.25 \text{ eV}

The minimum energy required to excite a hydrogen atom from its ground state ($n=1$) to the first excited state ($n=2$) is $13.6 \text{ eV} \times \left(1 - \dfrac{1}{4}\right) = 10.2 \text{ eV}$.

Since the maximum available energy ($6.25 \text{ eV}$) is less than the minimum excitation energy ($10.2 \text{ eV}$), the hydrogen atom cannot be excited. Thus, the collision must be perfectly elastic.

The problem states that the neutron's line of motion remains unchanged, which implies a head-on (one-dimensional) collision. In a perfectly elastic head-on collision between two bodies of equal mass, the bodies exchange their velocities.

Since the hydrogen atom was initially at rest, the neutron will come to rest after the collision.

Therefore, the final kinetic energy of the neutron is zero.