Modern Physics 1 Question 24
21. The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the following statement is true?
(2000, 2M)
(a) Its kinetic energy increases and its potential and total energy decreases
(b) Its kinetic energy decreases, potential energy increases and its total energy remains the same
(c) Its kinetic and total energy decreases and its potential energy increases
(d) Its kinetic, potential and total energy decreases
Show Answer
Solution:
- $v _n \propto \frac{1}{n} \therefore KE \propto \frac{1}{n^{2}}$ (with positive sign)
Potential energy $U$ is negative and $U _n \propto \frac{1}{r _n}$
$$ \begin{gathered} U _n=-\frac{1}{4 \pi \varepsilon _0} \cdot \frac{Z e^{2}}{r _n} \\ E _n \propto \frac{1}{n^{2}} \quad \quad \quad \text { (because } r _n \propto n^{2} \text { ) } \end{gathered} $$
Similarly, total energy $E _n \propto \frac{1}{n^{2}}$. (with negative sign)
Therefore, when an electron jumps from some excited state to the ground state, value of $n$ will decrease. Therefore, kinetic energy will increase (with positive sign), potential energy and total energy will also increase but with negative sign.
Thus, finally kinetic energy will increase, while potential and total energies will decrease.
NOTE
- For hydrogen and hydrogen-like atoms
$$ \begin{gathered} E _n=-13.6 \frac{Z^{2}}{n^{2}} eV \\ U _n=2 E _n=-27.2 \frac{Z^{2}}{n^{2}} eV \text { and } K _n=\left|E _n\right|=13.6 \frac{Z^{2}}{n^{2}} eV \end{gathered} $$
From these three relations we can see that as $n$ decreases, $K _n$ will increase but $E _n$ and $U _n$ will decrease.
- As an electron comes closer to the nucleus, the electrostatic force (which provides the necessary centripetal force) increases or speed (or KE) of the electron increases.
