Question

The electron in a hydrogen atom makes a transition from an excited state to the ground state, which of the following statements is true?

• A. Its kinetic energy increases and its potential and total energy decreases.
• B. Its kinetic energy and potential energy increases and its total energy remains same.
• C. Its kinetic energy and total energy decreases and potential energy increases.
• D. Its kinetic energy, potential energy and total energy decreases.

Answer 1

The correct option is A Its kinetic energy increases and its potential and total energy decreases.

The kinetic energy of the electron in the $n^{th}$ orbit is given by,

$(KE{)}_n=\frac{1}{2}\frac{kZ{e}^2}{r_n}$

$\Rightarrow (KE{)}_n\propto \frac{1}{r_n}$

Given, electron transition from the higher orbit to the lower orbit, thus radius decreases.
Hence, its kinetic energy increases.

Similarly, potential energy of the electron in the $n^{th}$ orbit is given by,

$(PE{)}_n=-\frac{kZ{e}^2}{r_n}[\because \left|\begin{array}{c}(PE{)}_n\end{array}\right|=2(KE{)}_n=]$

$\Rightarrow (PE{)}_n\propto -\frac{1}{r_n}$

As $r_n$ decreases, $(PE{)}_n$ also decreases.

Now, total energy of the electron in the $n^{th}$ orbit is given by,

$TE=PE+KE=-\frac{kZ{e}^2}{r_n}+\frac{kZ{e}^2}{2r_n}=-\frac{kZ{e}^2}{2r_n}$

$(TE{)}_n=-\frac{kZ{e}^2}{r_n}[\because \left|\begin{array}{c}(PE{)}_n\end{array}\right|=2(TE{)}_n=2(KE{)}_n]$

$\Rightarrow (TE{)}_n\propto -\frac{1}{r_n}$

As $r_n$ decreases, $(TE{)}_n$ also decreases.

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

Why this question ? Key concept: The relation between kinetic, potential, and total energy of the $n^{th}$ state is given by, $TE=PE+KE=-\frac{kZ{e}^2}{r_n}+\frac{kZ{e}^2}{2r_n}=-\frac{kZ{e}^2}{2r_n}$ $(TE{)}_n=-\frac{kZ{e}^2}{r_n}[\because \left|\begin{array}{c}(PE{)}_n\end{array}\right|=2(TE{)}_n=2(KE{)}_n]$