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Two H atoms in the ground state collide ...

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

A

10.2eV

B

20.4eV

C

13.6eV

D

27.2eV

Text Solution

Verified by Experts

The correct Answer is:
A
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`.
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