Two H atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is

Options

A. 10.2eV
B. 20.4eV
C. 13.6eV
D. 27.2eV

Answer

A

Explanation

In inelastic collision kinetic energy is not conseved so some part of K.E. is lost.
`therefore` Reduction in K.E.=
K.E. before collision -K.E. after collision
Now, since initial K.E. of each to two hydrogen atoms in ground state `=13.6eV`
`therefore` Total K.E. of both hydrogen atom beofore collision `=2xx13.6=27.2eV`
If one H atom goes over to first excited state `(n_1=2)` and other remains in ground state `(n_1=1)` then they combined K.E. after collision is `=(13.6)/((2)^2)+(13.6)/((1)^2)=3.4+13.6=17eV`
Hence reduction in K.E. `=27.2-17=10.2eV`.