Question

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

• A. 10.2 eV
• B. 20.4 Ev
• C. 13.6 eV
• D. 27.2 eV

Answer 1

Answer: (A)

10.2 eV

Answer 2

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 $= 13.6eV$

$\therefore$Total K.E. of both hydrogen atom before collision $= 2 \times 13.6 = 27.2eV$

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

$= \frac{13.6}{(2)^{2}} + \frac{13.6}{(1)^{2}} = 3.4 + 13.6 = 17eV$

Hence reduction in K.E. $= 27.2 - 17 = 10.2eV$