Question

The force of attraction between the positively charged nucleus and the electron in a hydrogen atom is given by $F=k\frac{e^2}{r^2}$. Assume that the nucleus is fixed. The electron, initially moving in an orbit of radius $R_1$ jumps into an orbit of smaller radius $R_2.$ The decrease in the total energy of the atom is-

• A. $\frac{k{e}^2}{2}(\frac{1}{R_1}+\frac{1}{R_2})$
• B. $\frac{k{e}^2}{2}(\frac{R_1}{R_2^2}-\frac{R_2}{R_1^2})$
• C. $\frac{k{e}^2}{2}(\frac{1}{R_2}-\frac{1}{R_1})$
• D. $\frac{k{e}^2}{2}(\frac{1}{R_1^2}-\frac{1}{R_2^2})$

Answer 1

The correct option is C $\frac{k{e}^2}{2}(\frac{1}{R_2}-\frac{1}{R_1})$

Given, the force between the nucleus and the revolving electron is,

$F=\frac{k{e}^2}{r^2}$

$\therefore \frac{k{e}^2}{r^2}=\frac{m{v}^2}{r}$

$\Rightarrow m{v}^2=\frac{k{e}^2}{r}$

$\Rightarrow \frac{1}{2}m{v}^2=\frac{k{e}^2}{2r}$

Total energy of the electron in a stationary orbit is,

$\text{T.E}=\text{K.E}+\text{P.E}$

$\text{T.E}=\frac{k{e}^2}{2r}+(-\frac{k{e}^2}{r})$

$\text{T.E}=(-\frac{k{e}^2}{2r})$

Also, we know

$\text{T.E}=-\text{K.E}=(-\frac{k{e}^2}{2r})$

Replace $r$ into $R$ in the above relation, we get,

$\text{T.E}=-\text{K.E}=(-\frac{k{e}^2}{2R})$

Let, total energy in the stationary orbit $R_1$ and $R_2$ be ${\text{T.E}}_i$ and ${\text{T.E}}_f$ respectively.

${\text{T.E}}_i=(-\frac{k{e}^2}{2R_1})$

${\text{T.E}}_f=(-\frac{k{e}^2}{2R_2})$

The decrease energy in the total energy is,

$\Delta E={\text{T.E}}_i-{\text{T.E}}_f=\frac{-k{e}^2}{2R_1}-\left(\frac{-k{e}^2}{2R_2}\right)$

$=\frac{k{e}^2}{2}(\frac{1}{R_2}-\frac{1}{R_1})$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.