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Two H atoms in the ground state collide ...

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

A

`10.20eV`

B

`20.40eV`

C

`13.6eV`

D

`27.2eV`

Text Solution

Verified by Experts

The correct Answer is:
A
Total mechanical energy of both the atoms before collision in the ground state,
`=2E=2(-13.6)=-27.eV`
`rArr` Total kinetic energy of electron in both the atoms
`2K=2(-E)=-2E=-(-27.2)=27.2eV ( :. K=-E)`
Now, after collision, total kinetic energy remaining with both the atoms is `(K_(1)+K_(2).)`. For this value to become maximum, amount of energy exchanged in the inelastic collision must be minimum. For this if electron does not do any transition in both the atoms then there will not be any decrease in total kinetic energy of two atoms. So this can not be accepted. There is another way.
Suppose electron in any one atom remains in ground state only. And electron in another atom transits from n = 1 to n = 2 by absorbing minimum amount of energy so that energy exchange will take place and that too with minimum amount. Here in this case total kinetic energy of both the atoms after collision will be
`=K_(1).+K_(2).`
`=-E_(1).+(-E_(2).)`
`=-(-13.6)-(-3.4)`
`( :. E_(2).=-(13.6)/(n^(2))=-(13.6)/(4)=-3.4eV)`
`=13.6+3.4=17eV`
Decrease in total kinetic energy of both the atoms,
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