Suppose energ $\Delta E$ is used this way and $v_1$ and $v_2$ are the speed of the neutron and hydrogen atom after collision.
According to law of conservation of linear momentum, $m v=m v_1+m v_2 \ldots \ldots$. (Equation $1)$
According to the law of conservation of energv, $\frac{1}{2} m v^2=\frac{1}{2} m v_1^2+\frac{1}{2} m v_2^2+\Delta E_{\ldots} \ldots \ldots$ (Equation $2)$
From equation 1, we get $v^2=v_1^2+v_2^2+2 v_1 v_2$
From equation 2, we get $v^2=v_1^2+v_2^2+2(\Delta E / m)$
$\therefore 2 v _1 v _2=2(\Delta E / m )$
$\therefore\left(v_1-v_2\right)^2=\left(v_1+v_2\right)^2+\left(-4 v_1 v_2\right)$
$=v^2-(4 \Delta E / m)$
As $\left(v_1-v_2\right)$ must be real, hence $\left[v^2-(4 \Delta E / m)\right] \geq 0$
or $\frac{1}{2} m v^2 \geq 2 \Delta E$
The minimum energythat can be absorbed bythe hydrogen atom in ground state to go in an excited is
$10.2\,eV$. Thus, minimum kinetic energy of the neutron needed for an inelastic collision
$=2 \times 10.2=20.4\,eV$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.