Question

A neutron moving with a speed υ strikes a hydrogen atom in ground state moving towards it with the same speed. Find the minimum speed of the neutron for which inelastic (completely or partially) collision may take place. The mass of neutron = mass of hydrogen = 1.67 × 10−27 kg.

Answer 1

The minimum speed of the neutron  with mass $1.67 \times 10^{-27}$ is  $6 \times 10^4$ meter per second for which inelastic (completely or partially) collision may takes place

Explanation:

It is given that  

Neutron’s mass, m = 1.67 × 10−27 kg

Since neutron is traversing with velocity v, its energy E is shown as

$E = 12 m v^2$

Let the absorbed energy be $\Delta E$.

For inelastic collision, the condition is shown as below:

$12mv^2 > 2\Delta E$

$\Delta E < 14mv^2$

For the first excited state, the required energy is$\Delta E < 10.2 eV$

Therefore, $14mv^2 > 10.2 eV$

So, the neutron’s minimum speed is shown as                    

$v_(min) = \sqrt{\frac{4\times 10.2 \ eV }{m} }$

$v_min = 6 \times 10^4$ meter per second

The minimum speed of the neutron is  $6 \times 10^4$ meter per second for which inelastic collision may takes place