A monochromatic beam of light having photon energy `12.5 eV` is incident on a simple `A` of atomic hydrogen gas in which all almost are in the ground state. The emission spectra obtained from this sample is incident on another sample `B` of atomic hydrogen gas in which all atoms are in the first excited state.
Based on above information, answer the following question:
The atoms of sample `A` after passing of light through it

Options

A. may be in the first excited state
B. may be in the second excited state
C. may be in both first and second excited state
D. None of above.

Explanation

To solve the problem, we need to analyze the situation step by step:

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

- For \( n = 1 \) (ground state):
\[ E_1 = -13.6 \, \text{eV} \]

- For \( n = 2 \) (first excited state):
\[ E_2 = -3.4 \, \text{eV} \]

- For \( n = 3 \) (second excited state):
\[ E_3 = -1.51 \, \text{eV} \]

- For \( n = 4 \) (third excited state):
\[ E_4 = -0.85 \, \text{eV} \]

### Step 2: Determine the Photon Energy
The photon energy incident on sample A is given as \( 12.5 \, \text{eV} \).

### Step 3: Calculate the Total Energy After Photon Absorption
When the photon of energy \( 12.5 \, \text{eV} \) hits the hydrogen atom in the ground state, the total energy becomes:
\[ E_{\text{final}} = E_1 + E_{\text{photon}} = -13.6 \, \text{eV} + 12.5 \, \text{eV} = -1.1 \, \text{eV} \]

### Step 4: Analyze the Final Energy Level
Now we need to determine if \( -1.1 \, \text{eV} \) corresponds to any of the quantized energy levels of hydrogen:

- The energy levels are:
- \( E_1 = -13.6 \, \text{eV} \)
- \( E_2 = -3.4 \, \text{eV} \)
- \( E_3 = -1.51 \, \text{eV} \)
- \( E_4 = -0.85 \, \text{eV} \)

Since \( -1.1 \, \text{eV} \) does not match any of the quantized energy levels, it means the atom cannot remain in a stable state at this energy.

### Step 5: Conclusion
Since the energy \( -1.1 \, \text{eV} \) does not correspond to any of the allowed energy levels, the atom of sample A cannot be in the first excited state, second excited state, or any other state. Therefore, after the photon passes through sample A, the atoms remain in the ground state.

### Final Answer
The atoms of sample A after passing through the light remain in the ground state.

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