Question

A hydrogen atom moves with a velocity $u$, and makes a head on perfectly inelastic collision with another stationary hydrogen atom. Both the hydrogen atom are in the ground state before collision. What is the minimum value of $u$, if one of them is to be given a minimum excitation energy?
The ionization energy of hydrogen atom is $13.6\text{eV}$. Mass of the hydrogen atom is $1.67\times {10}^{-27}\text{kg}$.

• A. $3.12\times {10}^4{\text{ms}}^{-1}$
• B. $6.24\times {10}^4{\text{ms}}^{-1}$
• C. $12.48\times {10}^4{\text{ms}}^{-1}$
• D. $9\times {10}^4{\text{ms}}^{-1}$

Answer 1

The correct option is B $6.24\times {10}^4{\text{ms}}^{-1}$

Let both the hydrogen atoms move with a velocity $v$ after collision.

From the conservation of momentum

$mu=2mv$

$\Rightarrow v=u/2$

The loss in $KE$ during collision is given by,

$\Delta E=\frac{1}{2}m{u}^2-2\times \frac{1}{2}m{\left(\frac{u}{2}\right)}^2$

$\Delta E=\frac{1}{4}m{u}^2........\left(1\right)$

This loss in $KE$ is utilized as the excitation energy.

The minimum excitation energy for a hydrogen atom is for transition $n_1=1$ to $n_2=2$, which corresponds to an energy of $10.2\text{eV}$.

$\Rightarrow \Delta E=10.2\text{eV}$

Substituting above value in $\left(1\right)$,

$10.2\times (1.6\times {10}^{-19})=\frac{1}{4}\times (1.67\times {10}^{-27})\times u^2$

$\Rightarrow \begin{array}{c}u=6.24\times {10}^4{\text{ms}}^{-1}\end{array}$

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