To understand why the correct answer is 6, we need to know about the energy levels of a hydrogen atom and the concept of spectral lines.

Energy levels of a hydrogen atom:

A hydrogen atom can exist in various energy levels, also known as orbitals. The lowest energy level is called the ground state, and all other levels are excited states. When an atom absorbs energy, its electrons move to higher energy levels, and when the electrons fall back to lower energy levels, they emit energy in the form of electromagnetic radiation.

Spectral lines:

Each element emits a unique set of wavelengths of electromagnetic radiation when its electrons move from higher energy levels to lower energy levels. These wavelengths of radiation create a pattern of lines in a spectrum, known as spectral lines. The number of spectral lines depends on the number of possible transitions between energy levels.

Now, let's apply this knowledge to the given problem:

The hydrogen atom in the ground state is excited by a monochromatic radiation of λ = 975 A. This means that the atom's electron has absorbed a specific amount of energy, causing it to move to a higher energy level. When the electron falls back to the ground state, it will emit energy in the form of electromagnetic radiation with a specific wavelength. This wavelength will create a spectral line in the spectrum.

The energy of the photon absorbed by the electron can be calculated using the equation E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength of the radiation. In this case, we get E = (6.626 x 10^-34 J s x 3 x 10^8 m/s) / (975 x 10^-10 m) = 2.03 x 10^-18 J.

This amount of energy is enough to excite the electron to the n = 2 energy level. When the electron falls back to the ground state (n = 1), it will emit energy in the form of electromagnetic radiation with a specific wavelength. The wavelengths of all possible transitions between energy levels in a hydrogen atom can be calculated using the Rydberg formula:

1/λ = R (1/n1^2 - 1/n2^2)

where λ is the wavelength of the emitted radiation, R is the Rydberg constant, and n1 and n2 are the initial and final energy levels, respectively.

For the hydrogen atom, the Rydberg constant is 1.097 x 10^7 m^-1.

Using the Rydberg formula, we can calculate the wavelengths of all possible transitions from the n = 2 energy level to the n = 1 energy level:

1/λ = R (1/2^2 - 1/1^2) = R/4

λ = 4/ R = 121.6 nm

This wavelength corresponds to the famous Lyman series of spectral lines in the ultraviolet region. However, this is not the only possible transition.

We can also calculate the wavelengths of all possible transitions from the n = 3 energy level to the n = 1 energy level:

1/λ = R (1/3^2 - 1/1^2) = 8R/9

λ = 9/8R = 102.6 nm

This wavelength corresponds to the Balmer series of spectral lines in the ultraviolet region.

Similarly, we can calculate the wavelengths of all