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

Answer 1

Answer: (B)

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 $=\Delta E$
From the law of conservation of linear momentum,
$mv=m{v}_1+m{v}_2$ ...(1)
And for law of conservation of energy,
$\frac{1}{2}m{v}^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2$ ...(2)
From squaring eq. (i), we get
$v^2=v_1^2+v_1^2+2v_1{v}_2$ ...(3)
From squaring eq. (ii), we get
$v_2=v_1^2+v_2^2+\frac{2\Delta E}{m}$
.(4) From eqn (3) & (4)
$\therefore 2v_1{v}_2=\frac{2\Delta E}{m}$
$\therefore {\left(v_1-v_2\right)}^2={\left(v_1+v_2\right)}^2-4v_1{v}_2$
$=v^2-\frac{4\Delta }{m}$
As, $v_1-v_2$ must be real, $v^2-\frac{4\Delta E}{m}\ge 0$
$\Rightarrow \frac{1}{2}m{v}^2\ge 2\Delta E$
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
$\frac{1}{2}m{v}_{\min }^2=2\times 10.2=20.4eV$