Travelling with a speed v , a neutron strikes a hydrogen atom in...

The neutron and the hydrogen atom have the same mass $m$ and move towards each other with the same speed $v$.

Since they have equal and opposite momenta, the center of mass of the system is at rest. The total kinetic energy of the system is available for the inelastic collision.

The total kinetic energy is given by:

$K = \dfrac{1}{2}mv^2 + \dfrac{1}{2}mv^2 = mv^2$

For an inelastic collision to occur, this kinetic energy must be at least equal to the minimum excitation energy of the hydrogen atom, which corresponds to the transition from the ground state ($n=1$) to the first excited state ($n=2$).

The excitation energy is:

$\Delta E = 13.6 \left( \dfrac{1}{1^2} - \dfrac{1}{2^2} \right) \text{ eV} = 10.2 \text{ eV}$

Equating the kinetic energy to the excitation energy:

$mv^2 = 10.2 \text{ eV}$

Converting the energy into Joules:

$10.2 \text{ eV} = 10.2 \times 1.6 \times 10^{-19} \text{ J} = 1.632 \times 10^{-18} \text{ J}$

Substituting the mass $m = 1.67 \times 10^{-27} \text{ kg}$:

$v^2 = \dfrac{1.632 \times 10^{-18}}{1.67 \times 10^{-27}} \approx 9.772 \times 10^8 \text{ m}^2/\text{s}^2$

$v = \sqrt{9.772 \times 10^8} \approx 3.126 \times 10^4 \text{ m/s}$

Rounding off, we get $v \approx 3.13 \times 10^4 \text{ m/s}$.