A hydrogen atom is excited from the ground state to a higher energy level with principal quantum number n = 4. How many distinct spectral lines will appear in the emission spectrum when the atom returns to the ground state through all possible transitions?

Introduction / Context:
The line spectrum of the hydrogen atom arises because electrons can occupy only discrete energy levels and emit photons when they jump from a higher level to a lower level. This question explores how many different spectral lines can result when an atom initially excited to a particular energy level makes all possible downward transitions. It tests both your understanding of quantum energy levels and your ability to apply a simple combinational formula.

Given Data / Assumptions:

• We have a hydrogen atom with one electron.
• The electron is excited from the ground state (n = 1) to a higher energy level with principal quantum number n = 4.
• The electron can return to the ground state through various intermediate energy levels, emitting photons at each transition.
• We assume all possible downward transitions occur at some stage.

Concept / Approach:
When an electron is at the energy level n, it can make transitions to any lower level (n - 1, n - 2 and so on down to 1). Each distinct pair of energy levels (n2, n1) with n2 > n1 corresponds to one possible spectral line, because the energy difference and therefore the wavelength of the emitted photon is unique. For a given upper level n, the total number of spectral lines that can be produced when the electron completes all possible transitions down to the ground level is given by the formula n * (n - 1) / 2. This counts all unique downward transitions between levels 1 to n.

Step-by-Step Solution:
Step 1: Identify the highest level that the electron reaches: n = 4. Step 2: Recall the formula for the number of possible spectral lines when all transitions from level n down to 1 are allowed: Number of lines = n * (n - 1) / 2. Step 3: Substitute n = 4 into the formula: Number of lines = 4 * (4 - 1) / 2. Step 4: Simplify: 4 * 3 / 2 = 12 / 2 = 6. Step 5: Therefore, a total of 6 distinct spectral lines can appear in the emission spectrum.

Verification / Alternative check:
We can list all downward transitions explicitly to confirm the count. From n = 4, the electron can drop to n = 3, 2 or 1. From n = 3, it can further drop to n = 2 or 1. From n = 2, it can drop to n = 1. So the possible transitions are: 4 → 3, 4 → 2, 4 → 1, 3 → 2, 3 → 1 and 2 → 1. Counting these gives 6 transitions, each corresponding to one spectral line at a different wavelength. This matches the formula result and confirms that 6 is correct.

Why Other Options Are Wrong:
2: This would be the number of transitions if only a couple of specific jumps were allowed, but here all possible transitions are considered.

5: This is one less than the correct count and might arise from forgetting one of the possible transitions, such as 4 → 2 or 3 → 1.
3: This could be guessed by taking n - 1 instead of n * (n - 1) / 2, but it does not count all pairs of levels.

Common Pitfalls:
Students often confuse the number of possible final levels (which is n - 1) with the number of distinct transitions. Another mistake is counting only transitions that end at the ground state and ignoring intermediate transitions between excited levels. Remember that each drop from any higher level to any lower level generates a spectral line. Using the formula n * (n - 1) / 2 is a reliable shortcut, and listing transitions explicitly is a good way to verify your answer in small cases.

Final Answer:
The number of spectral lines obtained when the hydrogen atom drops from n = 4 back to the ground state through all possible paths is 6.