Atoms — Physics STD 12 Science — Question

1. 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:

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.