Since it's mentioned that only one H -atom is present, (which means there is only one electron) the max. no of lines of different wavelengths that can be emitted from 7th EXCITED state (n=8) is =n-1 =7 lines.(8 to 7, 7 to 6....)
But if they had asked about a H-sample.... (which has large no. of electrons) .... the no. of different lines emitted will be 8×(8-1)/2 = 28. (which includes all possible transitions)

Maximum No. of Wavelengths Emitted by De-Excitation of Hydrogen Atom from 7th Excited State

The de-excitation of a hydrogen atom from its 7th excited state involves the electron transitioning from the 7th energy level to a lower energy level. During this process, energy is released in the form of photons, which can be calculated to determine the maximum number of wavelengths emitted.

Calculation:
- The energy levels in hydrogen are given by the formula E = -13.6/n^2 eV, where n is the principal quantum number.
- For the 7th excited state, n = 7, so the energy is -13.6/7^2 = -13.6/49 eV.
- The energy difference between the 7th and 1st energy levels is 13.6 eV, which corresponds to the energy of a photon emitted during de-excitation.
- The energy of a photon is given by E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon.
- By substituting the values, we can calculate the wavelength of the emitted photon.
- The maximum number of wavelengths emitted can be found by dividing the energy difference by the energy of one photon.

Result:
- By performing the calculations, we can determine the maximum number of wavelengths emitted by the de-excitation of a hydrogen atom from its 7th excited state.

This detailed explanation illustrates the process of calculating the maximum number of wavelengths emitted during the de-excitation of a hydrogen atom from its 7th excited state.