Question

How much energy (in eV) should be given to an electron in the level n = 2 of a hydrogen atom, so that it emit the Hβ-line of the Balmer series? R = 1.097 X 10⁷ m⁻¹, c = 3*10⁸ ms⁻¹, h = 6.625*10⁻³⁴ Js. [Ans: 2.56 eV]

Answer 1

With given ni= to nf=6.

With the support of the Rydberg equation and 1/1/λ= R( 1/n2-1/n2i) and here λ the wavelength of the photon , R is the Rydberg constant and equal to 1.097.107m-1.

With the given values 1/λ=1.097.107 m-1.(1/22-1/62) and 1/λ=2.4378.106 m-1 and the λ=4.10.10-7 m and hence the when the electron falls from ni=6 to nf=2 and photon of wavelength 410nm.

Hence it is photon of same wavelength in the fine manner.