I able to find the wavelength. But how to find which series it belongs to?

Solution:- 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. Also, gaseous hydrogen in its ground state at RT = -13.6 eV

So, the energy of the gaseous hydrogen = -13.6 + 12.5 = -1.1 eV

So, orbital energy E = -13.6/ n2 eV  (where, n = 3)

E = -13.6/ 32 = -13.6/9 = -1.5 eV

If the energy is equal to the energy of gaseous hydrogen. It can be concluded that the electron has jumped from n =1 to n - 3 level

In de-excitation process, electrons can jump from n=3 to n=1. then Lyman series as

1/ λ = Ry (1/ 12-1/ n2) , where, Ry =  1.097 x 107 m/s

Wavelength of radiation emitted by transition of electron

For n=3,

We can get

1/ λ = 1.097 x 107 (1/ 12-1/ 32) = 1.097 x 107(1- 1/9) = 1.097 x 107x 8/9 = 102.55 nm

If the transmission from n=2 to n=1 , then,

1/ λ = 1.097 x 107 (1/ 12-1/ 32) = 1.097 x 107(1- 1/4) = 1.097 x 107x 3/4 = 121.54 nm

If the transmission from n=3 to n=2 , then,

1/ λ = 1.097 x 107 (1/ 22-1/ 32) = 1.097 x 107 (1/4- 1/9) = 1.097 x 107x 5/36 = 656.33 nm

Hence, in Lyman series, two wavelengths i.e. 102.55 nm and 121.54 nm are emitted. In Balmer series, one wavelength, 656.33 nm is emitted.

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