Question: Q. 3. Two $H$ atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is (a) $10.20 \mathrm{eV}$. (b) $20.40 \mathrm{eV}$. (c) $13.6 \mathrm{eV}$. (d) $27.2 \mathrm{eV}$.
[NCERT Exemplar]
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Solution:
Ans. Correct option : (a)
Explanation : Total energy of two $\mathrm{H}$-atom in ground state $=2(-13.6)=-27.2 \mathrm{eV}$.
The maximum amount by which their combined kinetic energy is reduced when any one $\mathrm{H}$-atom goes into first excited state after the inelastic collision, that is, the total energy of two H-atom after inelastic collision :
$$ \begin{aligned} E & =\frac{13.6}{n^{2}}+13.6 \ & =\frac{13.6}{2^{2}}+13.61[\text { For excited stater }(n=2)] \ & =3.4+13.6=17.0 \mathrm{eV} \end{aligned} $$
So that the loss in kinetic energy due to inelastic collision will be,
$$ =27.2-17.0=10.2 \mathrm{eV} $$
