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Step-by-Step Solution
Step 1: Express the Initial Momentum and Final Momentum
A neutron of mass $m$ moving with speed $v$ makes a head-on collision with a stationary hydrogen atom (also assumed to have mass $m$ for simplicity). By conservation of momentum:
$mv = (m + m)\,v_{1}$
Hence, the common speed $v_{1}$ immediately after the inelastic collision is:
$v_{1} = \frac{v}{2}$
Step 2: Determine the Loss in Kinetic Energy
The initial kinetic energy of the neutron is
$ \frac{1}{2} m v^{2} $.
After the collision, both the neutron and the hydrogen atom move together with speed $v_{1} = \frac{v}{2}$. Therefore, the final kinetic energy of the combined mass $(2m)$ is:
$\frac{1}{2} \times (2m) \times \left(\frac{v}{2}\right)^{2}
= \frac{1}{2} \times 2m \times \frac{v^2}{4}
= \frac{1}{4}\, m v^{2}$
Thus, the loss in kinetic energy is:
$\frac{1}{2} m v^{2} - \frac{1}{4} m v^{2} = \frac{1}{4} m v^{2}$
Step 3: Relate the Lost Kinetic Energy to the Excitation Energy
This lost kinetic energy is used to excite the electron in the hydrogen atom from the ground state to the first excited state. The energy difference between these states is:
$E_{\text{excit}} = \bigl(-3.4\,\text{eV}\bigr) - \bigl(-13.6\,\text{eV}\bigr) = 10.2\,\text{eV}$
Hence, we have:
$\frac{1}{4}\,m v^{2} = 10.2\,\text{eV}$
Step 4: Find the Minimum Kinetic Energy of the Neutron
The initial kinetic energy of the neutron is $\frac{1}{2} m v^{2}$. From the relation above,
$\frac{1}{2}\,m v^{2} = 2 \times \frac{1}{4}\,m v^{2} = 2 \times 10.2\,\text{eV} = 20.4\,\text{eV}$
Therefore, the minimum kinetic energy of the neutron required for the inelastic collision to occur is:
$\boxed{20.4\,\text{eV}}$
