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Q.
Two $H$ atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is
Atoms
Solution:
In inelastic collision kinetic energy is not conserved 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 of two hydrogen atoms in
ground state $= 13.6\, eV$
$\therefore $ Total $K.E.$ of both Hydrogen atom before collision
$ = 2 \times 13.6 = 27.2 \,eV$
If one $H$ atom goes over to first excited state $(n_1= 2)$
and other remains in ground state $(n_2 = 1)$ then their combined $K.E.$ after collision is
$ = \frac {13.6}{(2)^2} + \frac{13.6}{(1)^2}$
$ = 3.4 + 13.6 = 17\,ev$
Hence reduced in $K.E. = 27.2 -17 $
$= 10. 2\,eV$