Question

When the excited electron of a $H$ atom from $n =5$ drops to the ground state, the maximum number of emission lines observed are________

Answer 1

Correct Answer: 4

To determine the maximum number of emission lines observed when an electron in a hydrogen atom drops from an excited state ($n=5$) to the ground state ($n=1$), we use the formula that calculates the total number of emission lines as a result of an electron transition between levels:

Formula: \(\text{Number of emission lines} = \frac{n(n-1)}{2}\), where \(n\) is the principal quantum number of the initial excited state.

Steps:

1. The electron starts at \(n=5\) and can transition to levels \(n=4, 3, 2, 1\).
2. Using the formula, substitute \(n=5\):
   \(\frac{5(5-1)}{2} = \frac{5 \times 4}{2} = \frac{20}{2} = 10\)
3. This calculation shows that the maximum number of emission lines is 10.

Each line corresponds to an electron transition from a higher to a lower energy level. All possible transitions are:

• \(n=5\) to \(n=4, 3, 2, 1\)
• \(n=4\) to \(n=3, 2, 1\)
• \(n=3\) to \(n=2, 1\)
• \(n=2\) to \(n=1\)

The computed value falls outside the provided range (4, 4). However, this discrepancy suggests reviewing either the problem understanding or the given range, as our solution logically concludes at 10 emission lines for the scenario described.