Question

An electron makes a transition from an excited state to the ground state of a hydrogen like atom. Out of the following statements which one is correct?

• A. Kinetic energy, potential energy and total energy decreases
• B. Kinetic energy and total energy decreased but potential energy increases
• C. Kinetic energy increases but potential energy and total energy decreases
• D. Kinetic energy decreases, potential energy increases but total energy remains the same.

Hint:

Remember the classic Bohr orbit relationships: $T.E. = -K.E.$ and $P.E. = 2 \times T.E.$. As an electron falls closer to the nucleus, it loses total energy (it becomes more negative/stable) and loses potential energy, while speeding up to maintain its orbit, thereby increasing its kinetic energy.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks about the behavior of kinetic energy (K.E.), potential energy (P.E.), and total energy (T.E.) when an electron drops from a higher energy level (excited state) down to a lower energy level (ground state) in a hydrogen-like atom.
When an electron makes a transition closer to the nucleus, the radius of its orbit decreases.

Step 2: Key Formula or Approach:
For a hydrogen-like atom with atomic number $Z$, the energies of an electron at an orbital radius $r$ are given by electrostatic relations:
Potential Energy: $$P.E. = -\frac{1}{4\pi\varepsilon_0} \frac{Ze^2}{r}$$ Kinetic Energy: $$K.E. = \frac{1}{8\pi\varepsilon_0} \frac{Ze^2}{r}$$ Total Energy: $$T.E. = K.E. + P.E. = -\frac{1}{8\pi\varepsilon_0} \frac{Ze^2}{r}$$

Step 3: Detailed Explanation:
When transitioning from an excited state to the ground state, the principal quantum number $n$ decreases, meaning the orbital radius $r$ decreases ($r \propto n^2$).
Let's look at how each energy term behaves as $r$ becomes smaller:
1. Kinetic Energy ($K.E. \propto \frac{1}{r}$): As $r$ decreases, the denominator decreases, causing the magnitude of $K.E.$ to increase. Since $K.E.$ is purely positive, its value strictly increases.
2. Potential Energy ($P.E. \propto -\frac{1}{r}$): As $r$ decreases, the fraction $\frac{1}{r}$ increases. However, due to the negative sign, a larger magnitude means a more negative value. Therefore, $P.E.$ decreases.
3. Total Energy ($T.E. \propto -\frac{1}{r}$): Just like potential energy, total energy contains a negative sign. As $r$ decreases, the value becomes more negative, meaning $T.E.$ decreases (energy is released as a photon).
Thus, $K.E.$ increases while $P.E.$ and $T.E.$ decrease.

Step 4: Final Answer:
The correct statement is that kinetic energy increases but potential energy and total energy decreases, which corresponds to option (C).