SATHEE: atoms Question 3

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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) $10.20 eV$ . (b) $20.40 eV$ . (c) $13.6 eV$ . (d) $27.2 eV$ .

[NCERT Exemplar]

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Solution:

Ans. Correct option : (a)

Explanation : Total energy of two $H$ -atom in ground state $= 2 (- 13.6) = - 27.2 eV$ .

The maximum amount by which their combined kinetic energy is reduced when any one $H$ -atom goes into first excited state after the inelastic collision, that is, the total energy of two H-atom after inelastic collision :

$\begin{aligned} E & = \frac{13.6}{n^{2}} + 13.6 & = \frac{13.6}{2^{2}} + 13.61 [For excited stater (n = 2)] & = 3.4 + 13.6 = 17.0 eV \end{aligned}$

So that the loss in kinetic energy due to inelastic collision will be,

$= 27.2 - 17.0 = 10.2 eV$

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