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 $= 13.6\, eV$ $\therefore$ 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 $K.E. = 27.2 -17$ $= 10. 2\,eV$