Question

A neutron moving with a speed ${}^{\prime }{v}^{\prime }$ makes a head on collision with a stationary hydrogen atom in ground state. The minimum kinetic energy of the neutron for which inelastic collision will take place is:

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

Answer 1

The correct option is A $20.4eV$

$n\to v\left(H\right)$ Before
$\left(n\right)\left(H\right)\to v/2$ After
loss in $K.E.=\frac{1}{2}m{v}^2-\frac{1}{2}\left(2m\right)(v/2{)}^2$
$=\frac{1}{4}m{v}^2$
K.E. lost is used to jump from 1st orbit to 2nd orbit $\delta K.E.=10.2ev$
$\Rightarrow \frac{1}{4}m{v}^2=10.2$
$\frac{1}{2}m{v}^2=2\times 10.2=20.4eV$