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,
$m v=m v_{1}+m v_{2}$ ...(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
$v^{2}=v_{1}^{2}+v_{1}^{2}+2 v_{1} v_{2}$ ...(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)
$\therefore 2 v_{1} v_{2}=\frac{2 \Delta E}{m}$
$\therefore \left(v_{1}-v_{2}\right)^{2} =\left(v_{1}+v_{2}\right)^{2}-4 v_{1} v_{2}$
$=v^{2}-\frac{4 \Delta}{m}$
As, $v_{1}-v_{2}$ must be real, $v^{2}-\frac{4 \Delta E}{m} \geq 0$
$\Rightarrow \frac{1}{2} m v^{2} \geq 2 \Delta E$
The minimum energy that can be absorbed by the hydrogen atom in the ground state to go into the excited state is $10.2\, eV$.
Therefore, the maximum kinetic energy needed is
$\frac{1}{2} m v_{\min }^{2}=2 \times 10.2=20.4\, eV$