Answer :

As we know, Energy in the ground state (n=1) is E where Z is atomic number so for energy of two H-atoms in the ground state, whose atomic number is Z=1, the energy will be twice , i.e..

After the inelastic collision between two H-atoms, one of the H-atom should go into first excited state (n=2) so that maximum amount of combined kinetic energy can be reduced. In other words the total energy of two H-atoms after inelastic collision will be

E=Ground state energy + Excited state energy

E [∵ for H-atom Z=1]

E [∵ for 1^{st} excited state n=2]

E

Now loss in K.E due to inelastic collision will be the maximum amount of combined kinetic energy reduced

i.e.

Therefore option (a) is correct.

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