In a sample of hydrogen like atoms all of which are in ground state, a photon beam containing photos of various energies is passed. In absorption spectrum, five dark line are observed. The number of bright lines in the emission spectrum will be (assume that all transitions take place)

Options

A. `21`
B. `10`
C. `15`
D. none of these

Explanation

To solve the problem, we need to analyze the information provided regarding the hydrogen-like atom and the observed absorption spectrum.

### Step-by-Step Solution:

1. **Understanding the Absorption Spectrum**:
- The problem states that five dark lines are observed in the absorption spectrum. Each dark line corresponds to a transition from the ground state (n=1) to an excited state.

2. **Determining the Final Quantum State**:
- Since there are five dark lines, it implies that there are five possible transitions to different excited states. The transitions can be represented as follows:
- n=1 to n=2
- n=1 to n=3
- n=1 to n=4
- n=1 to n=5
- n=1 to n=6
- Therefore, the highest excited state reached is n=6.

3. **Counting the Energy Levels**:
- The energy levels involved are n=1 (ground state) to n=2, n=3, n=4, n=5, and n=6 (excited states). Thus, the total number of energy levels is 6 (n=1 to n=6).

4. **Calculating the Number of Bright Lines in Emission Spectrum**:
- In the emission spectrum, bright lines correspond to transitions between excited states. The number of possible transitions between n=6 (highest excited state) and lower states can be calculated using the combination formula:
- The number of ways to choose 2 levels from 6 is given by \( C(6, 2) \).
- The formula for combinations is:
\[ C(n, r) = \frac{n!}{r!(n-r)!} \]
- For our case:
\[ C(6, 2) = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \]

5. **Conclusion**:
- Therefore, the number of bright lines in the emission spectrum is 15.

### Final Answer:
The number of bright lines in the emission spectrum will be **15**.