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 atom , such that one of the atoms reaches one of the excitation state.

Explanation

Let m be the mass of one hydrogen atom and u be the speed before collision. In case of inelastic collision both the particles will be moving together with the same speed and let this speed after collision be v.
As per conservation of linear momentum we can write the following equation :
mu+0=(m+m)v `rArr` v=u/2
Loss in K.E. due to collision can be calculated as follows :
`DeltaK_"loss"=K_1-K_2=((mu^2)/2)-1/2(2m)(u/2)^2 =(mu^2)/4`
Energy of hydrogen atom in `n^(th)` state is given by the following relation :
`E_n=-13.6/n^2` eV
Energy required to make the transition from n=1 to n=2 can be written as :
`DeltaE=-13.6 [1/2^2-1/1^2]=3/4xx13.6`=10.2 eV
This energy should be equal to the lost kinetic energy in collision.
`(m u^2)/4=10.2xx1.6xx10^(-19)`
Mass of hydrogen atom is `m=1.67xx10^(-27)` kg
`rArr u^2=(4xx10.2xx1.6xx10^(-19))/(1.67xx10^(-27))=39.09xx10^8`
`rArr u=6.25xx10^4` m/s