Question

A green dot in the middle of 4 concentric circles. The first away from the green dot is labeled n = 1. The second circle away from the green dot is labeled n = 2. The third circle away from the green dot is labeled n = 3 and the last is labeled n = 4. There is an arrow from n = 4 to n = 1 labeled A. There is an arrow from n = 3 to n = 2 labeled B. There is an arrow from n = 3 to n = 1 labeled C. There is an arrow from n = 2 to n = 1 labeled D.
Which transition by an electron will release the greatest amount of energy?

A
B
C
D

Answer 1

A

Energy released equals the difference in energy levels. Using E_n = −13.6 eV / n^2:
- n=4 → n=1: ΔE = 13.6(1 − 1/16) = 12.75 eV
- n=3 → n=1: ΔE = 13.6(1 − 1/9) = 12.09 eV
- n=2 → n=1: ΔE = 13.6(1 − 1/4) = 10.20 eV
- n=3 → n=2: ΔE = 13.6(1/4 − 1/9) = 1.89 eV

So transition A (n=4 → n=1) releases the most energy.