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 `B`

Options

A. will ionize when emission spectra of `A` is incident on `B`
B. may ionize when emission spectra of `A` is incident on `B`
C. will be excited to some higher state but won't ionize
D. None of above.

Explanation

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

### Step 1: Understand the Photon Energy and Atomic States
The photon energy incident on sample A is given as 12.5 eV. In atomic hydrogen, the energy levels are quantized, and the energy of the nth level is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
For hydrogen:
- Ground state (n=1): \( E_1 = -13.6 \, \text{eV} \)
- First excited state (n=2): \( E_2 = -3.4 \, \text{eV} \)
- Second excited state (n=3): \( E_3 = -1.51 \, \text{eV} \)
- Third excited state (n=4): \( E_4 = -0.85 \, \text{eV} \)

### Step 2: Determine the Effect of Incident Photon Energy on Sample A
The energy of the photon (12.5 eV) is significantly higher than the energy difference between the ground state and the ionization threshold (0 eV). Therefore, when the photon is absorbed by an atom in sample A, it can excite the atom to a higher energy state or ionize it.

Calculating the total energy after absorbing the photon:
\[ E_{\text{total}} = E_1 + 12.5 \, \text{eV} = -13.6 \, \text{eV} + 12.5 \, \text{eV} = -1.1 \, \text{eV} \]
Since -1.1 eV is still below 0 eV, the atom will not be ionized but will be in a state where it cannot exist (between n=2 and n=3). Thus, it will ionize.

### Step 3: Analyze the Emission Spectra from Sample A
When the hydrogen atoms in sample A are ionized, they will not emit any photons since they are no longer in bound states. Therefore, the emission spectra obtained from sample A will not have any photons to be incident on sample B.

### Step 4: Determine the State of Atoms in Sample B
Sample B consists of hydrogen atoms that are all in the first excited state (n=2). Since there is no emission spectra from sample A, the atoms in sample B will not receive any energy and will not be excited to higher states. They will eventually return to the ground state without any external influence.

### Conclusion
Based on the above analysis, the correct answer is that the atoms of sample B will not ionize, will not be excited to a higher state, and will simply return to the ground state.

### Final Answer
The atoms of sample B will **not ionize** and will **not be excited to a higher state**.

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