Question

A neutron moving with speed $v$ makes a head-on collision with a hydrogen atom in ground state kept at rest. Find the minimum kinetic energy of the neutron for which inelastic (completely or partially) collision may take place. The mass of neutron $\approx$ mass of hydrogen $=1.67\times {10}^{-27}kg$.

• A. $22.4eV$
• B. $21.4eV$
• C. $20.4eV$
• D. $20.1eV$

Answer 1

The correct option is C $20.4eV$

$\to$ Collision is inelastic if the whole or a part of the final kinetic energy is used to excite the hydrogen atom from ground to 1st exited state

1. Conservation of linear momentum,

$mv=m{v}_1+m{v}_2$----------------(1)

2. Conservation of energy,

$\frac{m{v}^2}{2}=\frac{m{v}_1^2}{2}+\frac{m{v}_2^2}{2}+\Delta \in$ ------(2)

Where, $\Delta \in$ is the excitation energy

$\to$ Solving 1 and 2 simultaneously,

$v_1^2-v{v}_1+\frac{2\Delta \in }{m}=0$

$\to$ For $v_1$ to be real,

$v^2-\frac{4\Delta \in }{m}\ge 0$

$\Rightarrow \frac{m{v}^2}{2}\ge 2\Delta \in$

$\to$ so minimum KE is $20.4eV$

Calculating $\Delta \in$....

$\to$ For hydrogen like atom,

$E_n=-z^2.\frac{13.61eV}{n^2}$

$E_2=-1^2.\frac{13.61eV}{2^2}$

$=-3.40eV$

$E_1=-13.61eV$

$\to \Delta \in =E_2-E_1=10.2eV$