Question

A moving hydrogen atom makes a head on collision with a stationary hydrogen atom. Before collision both atoms are in ground state and after collision they move together. What is the minimum value of the kinetic energy of the moving hydrogen atom, such that one of the atoms reaches one of the excitation state.

Answer 1

Let u be velocity of hydrogen atom before collision and V the ve locity of two atoms moving together after collision By principal of conservation of momentum , we have

$M_u+M\times 0=2MV$

$V=\frac{u}{2}$

The loss in kinetic energy $\Delta ϵ$ due to collision is

$\Delta ϵ=\frac{1}{2}M{u}^2-\frac{1}{2}\left(2M\right)V^2$

As $V=\frac{u}{2}$, we have $\Delta ϵ=\frac{1}{2}M{u}^2-\frac{1}{2}\left(2M\right){\left(\frac{u}{2}\right)}^2$

$=\frac{1}{2}M{u}^2-\frac{1}{4}M{u}^2=\frac{1}{4}M{u}^2$

This loss in energy is due to excitation of one of hydrogen atoms . The ground state $(n=1)$ energy of a hydrogen atom is ${ϵ}_1=-13.6$eV

The energy of first excited level $(n=2)$ is :${ϵ}_2=-3.4$eV

Thus minimum energy required to excite a hydrogen atom from ground state to first excited state is :

${ϵ}_2-{ϵ}_1=[-3.4-\left(13.6\right)]$eV$=10.2$eV

$=10.2\times 1.6\times {10}^{-19}$J$=16.32$J

As per problem with the loss in kinetic energy in collision is due to energy used up in exciting one of atoms

Thus, $\Delta ϵ={ϵ}_2-{ϵ}_1$

or $\frac{1}{4}M{u}^2=16.32\times {10}^{-19}$

or $u^2=\frac{4\times 16.32\times {10}^{-19}}{M}$

The mass of hydrogen atom is $1.0078$amu or $1.0078\times 1.66\times {10}^{-27}$kg

$u^2=\frac{4\times 16.32\times {10}^{-19}}{1.0078\times 1.66\times {10}^{-27}}=39.02\times {10}^8$kg

or $u=\sqrt{39.02\times {10}^8}=6.24\times {10}^4{ms}^{-1}$