Two H atoms in the ground state collide inelastically. The maximu…

MCQ

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

✓
$10.20\ eV.$
B
$20.40\ eV.$
C
$13.6\ eV.$
D
$27.2\ eV.$

✓

Answer

Correct option: A.

$10.20\ eV.$

Key concept: Total energy $(E)$ is the sum of potential energy and kinetic energy, i.e.$ E = K + U$
$\Rightarrow\ \text{E}=-\frac{\text{kZe}^2}{2\text{r}_\text{n}}\text{also r}_\text{n}=\frac{\text{n}^2\text{h}^2\in_0}{\pi\text{mze}^2}$
Hence $\text{E}=-\bigg(\frac{\text{me}^4}{8\in_0^2\text{h}^2}\bigg)\frac{\text{Z}^2}{\text{n}^2}=-\bigg(\frac{\text{me}^4}{8\in_0^2\text{ch}^3}\bigg)\text{ch}\frac{\text{Z}^2}{\text{n}^2}$
$=-\text{R ch}\frac{\text{Z}^2}{\text{n}^2}=-13.6\frac{\text{Z}^2}{\text{n}^2}\text{eV}$
The lowest state of the atom, called the ground state, is that of the lowest energy. The energy of this state $(n = 1), E_1$ is $-13.6\ eV$
Energy level diagram of hydrogen/hydrogen like atom:

Principal quantum number	Orbit	Excited state	Energy of $H_2 -$ atom
$\text{n}=\infty$	Infinite	Infinite	$0\ eV$
$n = 4$	Fourth	Third	$-0.85\ eV$
$n = 3$	Third	Second	$-1.51\ eV$
$n = 2$	Second	First	$-3.4\ eV$
$n = 1$	First	Ground	$-13.6\ eV$

Let two $H$ atoms initially at in the ground state.Now two stoms collide inclastically. The total energy associated with the tow $H-$atoms
$= 2 \times (13.6 eV) = 27.2\ eV$
The maximum amount by which their combined kinetic energy is reduced when any one of them goes into first excited state $(n = 2)$ after the inclastic collision.
The total energy associated with the two $H-$atoms after the collision
$=\Big(\frac{13.6}{2^2}\Big)+(13.6)=17.0\text{eV}$
Hence, maximum loss of their combined kinetic energy
$= 27.2 - 17.0 = 10.2\ eV.$