Question

A neutron moving with speed $v$ makes a head on collision with a hydrogen atom in ground state kept at rest. The minimum kinetic energy of the neutron for which inelastic collision takes place is

• A. $10.2eV$
• B. $20.4eV$
• C. $12.1eV$
• D. $16.8eV$

Answer 1

Answer: (B)

$20.4eV$

Let speed of neutron before collision $=V$
Speed of neutron after collision $=V_1$
Speed of proton or hydrogen atom after collision $=V_2$
Energy of excitation $=△E$
From the law of conservation of linear momentum,
$mv=m{v}_1+m{v}_2.....\left(1\right)$
And for law of conservation of energy,
$\frac{1}{2}m{v}^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2....\left(2\right)$
From squaring eq. (i), we get
$v^2=v_1^2+v_2^2+2v_1{v}_2.....\left(3\right)$
From squaring eq. (ii), we get
$v^2=v_1^2+v_2^2+\frac{2△E}{m}.....\left(4\right)$
From eqn $\left(3\right)$ and $\left(4\right)$
$\therefore 2v_1{v}_2=\frac{2△E}{m}$
$\therefore (v_1-v_2{)}^2=(v_1+v_2{)}^2-4v_1{v}_2=v^2-\frac{4△E}{m}$
As, $v_1-v_2$ must be real, $v^2-\frac{4△E}{m}\ge 0$
$\Rightarrow \frac{1}{2}m{v}^2\ge 2△E$
The minimum energy that can be absorbed by the hydrogen atom in the ground state to go into the excited state is $10.2eV$. Therefore, the maximum kinetic energy needed is
$\frac{1}{2}m{v}_{min}^2=2\times 10.2=20.4eV$.