Ionization potential of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1 eV. the spectral lines emitted by hydrogen atoms according to Bohrs theory will bea)Oneb)Twoc)Threed)FourCorrect answer is option 'C'. Can you explain this answer?

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To understand why the correct answer is option 'C', let's look at the concept of ionization potential and the excitation of hydrogen atoms.

1. Ionization potential:
The ionization potential of an atom is the amount of energy required to remove an electron from its ground state, resulting in the formation of a positively charged ion. For a hydrogen atom, the ionization potential is 13.6 eV.

2. Excitation of hydrogen atoms:
When a hydrogen atom absorbs energy, such as through the interaction with monochromatic radiation (photons), its electron can be excited to a higher energy level. The energy of the absorbed photon must match the energy difference between the electron's initial and final energy states.

Now, let's analyze the scenario described in the question:

- The ionization potential of a hydrogen atom is 13.6 eV.
- The monochromatic radiation has a photon energy of 12.1 eV.

We need to determine the number of spectral lines emitted by the hydrogen atoms according to Bohr's theory.

1. Calculation of the energy difference:
To determine the energy difference between the ground state and the excited state of the hydrogen atom, we subtract the energy of the ground state from the energy of the excited state.

Energy difference = Energy of excited state - Energy of ground state
Energy difference = 12.1 eV - 0 eV
Energy difference = 12.1 eV

2. Identification of spectral lines:
According to Bohr's theory, the energy levels of hydrogen atoms are quantized, and the electron can only occupy specific energy states. When an electron transitions from a higher energy state to a lower energy state, it emits a photon with energy equal to the energy difference between the two states.

In this case, the electron is transitioning from the excited state to the ground state, which results in the emission of a photon. Since the energy difference is 12.1 eV, the emitted photon will have an energy of 12.1 eV.

Now, we need to consider the possible transitions that can occur between different energy states. The energy levels in a hydrogen atom are given by the formula:

Energy = -13.6 eV/n^2

where n is the principal quantum number.

3. Determination of possible transitions:
To find the possible transitions, we need to calculate the principal quantum numbers (n) for the energy levels involved in the transition.

For the ground state, n = 1:
Energy of ground state = -13.6 eV/1^2 = -13.6 eV

For the excited state, n is unknown:
Energy of excited state = -13.6 eV/n^2 = 12.1 eV

Rearranging the equation:
n^2 = 13.6 eV/12.1 eV
n^2 = 1.12
n ≈ 1.06

Since the principal quantum number (n) must be an integer, the only possible value for n in this case is 1.

4. Conclusion:
Based on the calculations, we can conclude that the electron in the hydrogen atom transitions from the excited state (n = 1.06) to the ground state (n = 1). This results in the emission of a photon with an energy of 12.1 eV.

Therefore, according to Bohr's theory, only one spectral line

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