Question

An electron transition takes place from excited state to ground state in hydrogen atom, then

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

Hint:

The closer the electron is to the nucleus (lower \( n \)), the faster it must move to stay in orbit (higher kinetic energy), and the more tightly bound it is (lower/more negative potential and total energy).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The goal is to determine how the kinetic, potential, and total energy of an electron change when it moves from a higher energy level (excited state) to a lower energy level (ground state).
Step 2: Key Formula or Approach:
For an electron in the \( n \)-th orbit of a hydrogen atom:
- Total Energy: \( E_n = -13.6 \frac{1}{n^2} \text{ eV} \)
- Kinetic Energy: \( K_n = -E_n = 13.6 \frac{1}{n^2} \text{ eV} \)
- Potential Energy: \( U_n = 2E_n = -27.2 \frac{1}{n^2} \text{ eV} \)
Step 3: Detailed Explanation:
Transition is from an excited state (higher \( n \)) to the ground state (\( n = 1 \)).
1. Total Energy (\( E \)): As \( n \) decreases, \( \frac{1}{n^2} \) increases. Since \( E \) is negative, a larger numerical value means a decrease (it becomes more negative).
2. Kinetic Energy (\( K \)): \( K \) is positive and proportional to \( \frac{1}{n^2} \). As \( n \) decreases, \( K \) increases.
3. Potential Energy (\( U \)): Since \( U \) is negative and proportional to \( \frac{1}{n^2} \), as \( n \) decreases, \( U \) decreases (it becomes more negative).
Comparing with options, statement (1) is correct.
Step 4: Final Answer:
Upon transition to a lower state, kinetic energy increases while potential and total energy decrease. This is option (1).