Question

A neutron collides head-on with a stationary hydrogen atom in ground state. Which of the following statement(s) is / are correct?

$($Assume that mass of the neutron and mass of the hydrogen atom are equal$)$

• A. If the Kinetic energy of the neutron is less than $13.6\text{eV}$, collision must be elastic.
• B. If the Kinetic energy of the neutron is less than $13.6\text{eV}$, collision must be inelastic.
• C. Inelastic collision may take place on when the initial Kinetic energy of neutron is greater than $13.6\text{eV}$
• D. If the kinetic energy of the neutron is more than $20.4\text{eV}$, collision must be inelastic.

Answer 1

The correct option is D If the kinetic energy of the neutron is more than $20.4\text{eV}$, collision must be inelastic.

The collision will be inelastic if a part of the initial kinetic energy of the neutron is used to excite the atom.

Let $\Delta E$ be used to excite the atom, $v$ be the speed of the neutron before collision, $v_1$ and $v_2$ be the speed of the neutron and the speed of the hydrogen atom after collision.

By conservation of momentum,

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

By conservation of energy,

$\frac{1}{2}m{v}^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2+\Delta E---\left(2\right)$

From $\left(1\right)$, $v^2=v_1^2+v_2^2+2v_1{v}_2$

From $\left(2\right)$, $v^2=v_1^2+v_2^2+\frac{2\Delta E}{m}$

$\Rightarrow 2v_1{v}_2=\frac{2\Delta E}{m}---\left(3\right)$

We know that,

$(v_1-v_2{)}^2=(v_1+v_2{)}^2-4v_1{v}_2$

Substituting $\left(1\right)$ and $\left(3\right)$ in above equation,

$(v_1-v_2{)}^2=v^2-\frac{4\Delta E}{m}$

As, $(v_1-v_2)$ has to be real,

$\Rightarrow v^2>\frac{4\Delta E}{m}$

$\Rightarrow \frac{1}{2}m{v}^2>2\Delta E$

The minimum energy required to excite a hydrogen atom from ground state is $10.2\text{eV}$

Thus, the minimum kinetic energy of the neutron before collision required for inelastic collision should be $20.4\text{eV}$

Hence, option $\left(\text{D}\right)$ is correct.