Question

The ground state energy of hydrogen atom is −13.6 eV. If an electron make a transition from an energy level −0.85 eV to −1.51 eV, calculate the wavelength of the spectral line emitted. To which series of hydrogen spectrum does this wavelength belong?

Answer 1

Since energy of Hydrogen atom is given as

`E_n = (-13.6)/n^2  eV`

Ground state energy E₁ = −13.6eV

Now electron makes a transition from energy level

n_p = −0.85 eV to n_q = −1.51 eV

Now,

`E_(n_p) = (-13.6)/n_p^2  eV`

`n_p^2 = (-13.6)/-0.85 = 16`

`n_p = 4`

Again,`E_(n_p) = (-13.6)/n_q^2  eV`

          `n_p^2 = (-13.6)/(-1.51)  =9`

             `n_q = 3`

Thus, we have transition from n = 4 to n = 3.

Since transition corresponds to the transition of an electron from some higher energy state to an orbit having n = 3. It is Paschen series of the hydrogen spectrum. Wavelength is given as

`1/lambda = R(1/3^2 - 1/x^2)`

Where R is Rydberg’s constant

`R = 1.097 xx 10^7 m^-1 1/lambda = 1.097 xx 10^7 [1/9 - 1/16]`

  `= 1.097 xx 10^7 [7/(9 xx 16)]`

`lambda = (9 xx 16)/(1.097 xx 10^7 xx 7)  = 18.752 xx 19^-7 m = 1875 nm`