Question

The electron in hydrogen atom is initially in the third excited state. When it finally moves to ground state, the maximum number of spectral lines emitted are

• A. 3
• B. 4
• C. 5
• D. 6

Hint:

Always remember: the $n^{\text{th}}$ excited state means the actual energy level is $n + 1$. Once you have the correct $n$, just use the quick handshake formula $\frac{n(n-1)}{2}$ to get the number of lines instantly!

Answer 1

The Correct Option is D

The third excited state corresponds to the principal quantum number $n = 4$, while the ground state corresponds to $n = 1$. The maximum number of distinct spectral lines emitted when an electron de-excites from an energy level $n$ to the ground state ($n=1$) is given by the combination formula: $$N = \frac{n(n - 1)}{2}$$ Substituting $n = 4$ into the formula: $$N = \frac{4 \times (4 - 1)}{2} = \frac{12}{2} = 6$$ The 6 possible distinct transitions are: * From $n=4$: $4 \rightarrow 3$, $4 \rightarrow 2$, $4 \rightarrow 1$ * From $n=3$: $3 \rightarrow 2$, $3 \rightarrow 1$ * From $n=2$: $2 \rightarrow 1$
Final Answer:
The maximum number of spectral lines emitted is 6, which corresponds to option (D).