Question

X different wavelengths may be observed in the spectrum from a hydrogen sample if the atoms are exited to states with principal quantum number n=6? The value of X is _________.

Hint:

Memorizing the formula \(N = \frac{n(n-1)}{2}\) is the fastest way to solve this common type of problem. It's derived from the concept of combinations, as we are choosing any two energy levels out of n levels for a transition to occur.

Answer 1

Correct Answer: 15

Step 1: Understanding the Question:
We need to find the total number of possible spectral lines (corresponding to different wavelengths) emitted when an electron in a hydrogen atom de-excites from the n=6 energy level to lower energy levels.
Step 2: Key Formula or Approach:
When an electron is in the n-th excited state, it can make transitions to any of the lower states (n-1, n-2, ..., 1). The total number of possible spectral lines emitted is given by the formula:
\[ \text{Number of lines} = \frac{n(n-1)}{2} \] where n is the principal quantum number of the initial excited state.
Step 3: Detailed Explanation:
Given that the atoms are excited to the state with principal quantum number \(n=6\).
Using the formula, we can calculate the number of different wavelengths, X.
\[ X = \frac{6(6-1)}{2} \] \[ X = \frac{6 \times 5}{2} \] \[ X = \frac{30}{2} = 15 \] Alternatively, we can count the transitions:
- From n=6 to lower states (5, 4, 3, 2, 1): 5 lines
- From n=5 to lower states (4, 3, 2, 1): 4 lines
- From n=4 to lower states (3, 2, 1): 3 lines
- From n=3 to lower states (2, 1): 2 lines
- From n=2 to lower state (1): 1 line
Total lines = 5 + 4 + 3 + 2 + 1 = 15.
Step 4: Final Answer:
The value of X is 15.