Question

calculate the ratio of wavelength of line of minimum energy of lyman series and second line of paschen series of hydrogen atom​

Answer 1

Answer:

For Paschen Series, the formula for wavelength becomes: The value of n can be now 4,5,6,... We have to find the ratio of wavelength of first line to that of second line of Paschen Series. First line of Paschen Series is obtained by n=4.

The Lyman limit is the short-wavelength end of the hydrogen Lyman series, at 91.2 nm (912 Å). It corresponds to the energy required for an electron in the hydrogen ground state to escape from the electric potential barrier that originally confined it, thus creating a hydrogen ion.

When n = 3, Balmer's formula gives λ = 656.21 nanometres (1 nanometre = 10−9 metre), the wavelength of the line designated Hα, the first member of the series (in the red region of the spectrum), and when n = ∞, λ = 4/R, the series limit (in the ultraviolet).

Since wavelength and energy are inversely related, the longest wavelength would be produced by the lowest amount of energy. That would be by the electron falling from from n = 3 (ni) to n = 2 (nf). (All Balmer series lines have nf = 2). lambda = 1 / (1.52 x 10^6 m) = 6.56 x 10^-7 m = 656 nm . . .that's a RED line