In a sample of hydrogen atoms, all the atoms exist in two energy levels A and B. A is the ground level and B is some higher energy level. These atoms absorb photons of energy 2.7 eV and attain a higher energy level C.After this, these atoms emit photons of six different energies. Some of these photon energies are higher than 2.7 eV, some equal to 2.7 eV and some are loss than 2.7 eV.
The principal quantum number corresponding to energy level B is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to determine the principal quantum number corresponding to energy level B in a sample of hydrogen atoms. We know that:

1. **Energy Levels in Hydrogen**: The energy levels in a hydrogen atom are given by the formula:
\[ E_n = -\frac{13.6 \text{ eV}}{n^2} \]
where \( n \) is the principal quantum number.

2. **Photon Absorption**: The hydrogen atoms absorb photons of energy 2.7 eV to reach a higher energy level C.

3. **Photon Emission**: After reaching level C, the atoms emit photons of six different energies, which indicates multiple transitions between energy levels.

### Step-by-Step Solution:

**Step 1: Determine the number of transitions.**
The number of different photon energies emitted corresponds to the number of transitions an electron can make between energy levels. The formula for the number of transitions (or emissions) from an excited state with principal quantum number \( n \) is given by:
\[ \text{Number of emissions} = \frac{n(n-1)}{2} \]
Given that there are 6 different emissions, we set up the equation:
\[ \frac{n(n-1)}{2} = 6 \]
Multiplying both sides by 2 gives:
\[ n(n-1) = 12 \]

**Step 2: Solve the quadratic equation.**
Rearranging gives us:
\[ n^2 - n - 12 = 0 \]
Factoring this quadratic equation:
\[ (n - 4)(n + 3) = 0 \]
This gives us two solutions:
\[ n = 4 \quad \text{or} \quad n = -3 \]
Since \( n \) cannot be negative, we have:
\[ n = 4 \]

**Step 3: Identify the energy levels.**
We have established that the excited state C corresponds to \( n = 4 \). Now we need to find the principal quantum number for energy level B.

**Step 4: Determine the energy levels.**
Using the energy formula:
- For \( n = 1 \):
\[ E_1 = -\frac{13.6}{1^2} = -13.6 \text{ eV} \]
- For \( n = 2 \):
\[ E_2 = -\frac{13.6}{2^2} = -3.4 \text{ eV} \]
- For \( n = 3 \):
\[ E_3 = -\frac{13.6}{3^2} \approx -1.51 \text{ eV} \]
- For \( n = 4 \):
\[ E_4 = -\frac{13.6}{4^2} = -0.85 \text{ eV} \]

**Step 5: Calculate the energy difference.**
The energy difference between levels B and C must equal the absorbed energy of 2.7 eV:
- If \( B = 2 \) (ground state) and \( C = 4 \):
\[ E_C - E_B = (-0.85) - (-3.4) = 2.55 \text{ eV} \]
- If \( B = 1 \) and \( C = 4 \):
\[ E_C - E_B = (-0.85) - (-13.6) = 12.75 \text{ eV} \]
- If \( B = 3 \) and \( C = 4 \):
\[ E_C - E_B = (-0.85) - (-1.51) = 0.66 \text{ eV} \]

The closest match to the absorbed energy of 2.7 eV is when \( B = 2 \) and \( C = 4 \).

### Conclusion:
The principal quantum number corresponding to energy level B is:
\[ \boxed{2} \]