Question
Two $H$ atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is :
✓
Answer
$10.20eV.$
Key concept :
Total energy $(E)$ is the sum of potential energy and kinetic energy, i.e. $E = K + U$
$\Rightarrow\ \text{E}=-\frac{\text{kZe}^2}{2\text{r}_\text{n}}$ also ${r}_\text{n}=\frac{\text{n}^2\text{h}^2\epsilon_0}{\pi\text{mze}^2}$
Hence $\text{E}=-\bigg(\frac{\text{me}^4}{8\epsilon_0^2\text{h}^2}\bigg)\frac{\text{Z}^2}{\text{n}^2}=-\bigg(\frac{\text{me}^4}{8\epsilon_0^2\text{ch}^3}\bigg)\text{ch}\frac{\text{Z}^2}{\text{n}^2}$
$=-\text{R ch}\frac{\text{Z}^2}{\text{n}^2}=-13.6\frac{\text{Z}^2}{\text{n}^2}\text{eV}$
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 $-13.6eV$
Energy level diagram of hydrogen/hydrogen like atom:
Let two $H$ atoms initially at in the ground state.
Now two stoms collide inclastically.
The toal energy associated with the tow $H-$ atoms
$= 2 \times (13.6 eV) = 27.2eV$
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
$= 27.2 - 17.0 = 10.2eV.$
Key concept :
Total energy $(E)$ is the sum of potential energy and kinetic energy, i.e. $E = K + U$
$\Rightarrow\ \text{E}=-\frac{\text{kZe}^2}{2\text{r}_\text{n}}$ also ${r}_\text{n}=\frac{\text{n}^2\text{h}^2\epsilon_0}{\pi\text{mze}^2}$
Hence $\text{E}=-\bigg(\frac{\text{me}^4}{8\epsilon_0^2\text{h}^2}\bigg)\frac{\text{Z}^2}{\text{n}^2}=-\bigg(\frac{\text{me}^4}{8\epsilon_0^2\text{ch}^3}\bigg)\text{ch}\frac{\text{Z}^2}{\text{n}^2}$
$=-\text{R ch}\frac{\text{Z}^2}{\text{n}^2}=-13.6\frac{\text{Z}^2}{\text{n}^2}\text{eV}$
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 $-13.6eV$
Energy level diagram of hydrogen/hydrogen like atom:
| Principal quantum number | Orbit | Excited state | Energy of $H_2 -$ atom |
|
$\text{n}=\infty$
|
Infinite | Infinite | $0eV$ |
| $n = 4$ | Fourth | Third | $-0.85eV$ |
| $n = 3$ | Third | Second | $-1.51eV$ |
| $n = 2$ | Second | First | $-3.4eV$ |
| $n = 1$ | First | Ground | $-13.6eV$ |
Now two stoms collide inclastically.
The toal energy associated with the tow $H-$ atoms
$= 2 \times (13.6 eV) = 27.2eV$
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
$= 27.2 - 17.0 = 10.2eV.$
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