- 10.20eV.
Solution:
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), E1 is -13.6eV
Energy level diagram of hydrogen/hydrogen like atom:
| Principal quantum number | Orbit | Excited state | Energy of H2 - 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 |
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 × (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.
