A hydrogen sample is prepared in a particular excited state. Photons of energy `2.55 eV` get absorbed into the sample to take some of the electrons to a further excited state `B`. Find orbit numbers of the states `A` and `B`. Given the allowed energies of hydrogen atom:
`E_(1)=-13.6 eV, E_(2)=-3.4 eV, E_(3)=-1.5 eV, E_(4)=-0.85 eV. E_(6)=-0.54 eV`

Answer

To find the orbit numbers of states A and B for the hydrogen atom, we need to follow these steps:

### Step 1: Identify the energies of the hydrogen atom
The allowed energies for the hydrogen atom are given as:
- \( E_1 = -13.6 \, \text{eV} \)
- \( E_2 = -3.4 \, \text{eV} \)
- \( E_3 = -1.5 \, \text{eV} \)
- \( E_4 = -0.85 \, \text{eV} \)
- \( E_6 = -0.54 \, \text{eV} \)

### Step 2: Determine the energy difference
We know that photons of energy \( 2.55 \, \text{eV} \) are absorbed, which means we need to find two energy levels such that the difference between them equals \( 2.55 \, \text{eV} \).

### Step 3: Calculate the energy differences
We will calculate the differences between the allowed energy levels:

1. \( E_4 - E_2 = (-0.85) - (-3.4) = -0.85 + 3.4 = 2.55 \, \text{eV} \)

This shows that the transition occurs from \( E_2 \) to \( E_4 \).

### Step 4: Identify the orbit numbers
From the calculated energy levels:
- The initial state \( A \) corresponds to \( E_2 \), which is the second orbit (n=2).
- The final state \( B \) corresponds to \( E_4 \), which is the fourth orbit (n=4).

### Conclusion
Thus, the orbit numbers of states A and B are:
- State A (initial state) = 2
- State B (final state) = 4

### Final Answer
- Orbit number of state A: 2
- Orbit number of state B: 4

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