Question

The ground state energy of hydrogen atom is −13.6 eV. If and electron make a transition from the energy level −0.85 eV to −3.4 eV, calculate spectrum does his wavelength belong?

Answer 1

 `E_n = -(13.6)/n^2 ev.`Here ground state energy for n = 1, E₁ = −13.6eV

Now electron transits from E_p = −0.85 eV to E_q = −3.4 eV

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

`n_p^2 = 13.6/0.85 =16`

Thus,`n_p = 4`

Again,`-3.4 = (-13.6)/n_q^2`

`n_q^2 = 13.6/3.4 =4 `

`=> n_q = 2`

Thus electron makes transition from n = 4 to n = 2. Hence it is Balmer series. Now, R = 1.0974 × 10⁷ m⁻¹

`1/lambda = R (1/2^2 - 1/n^2) => 1/lambda = 1.0974 xx 10^7 (1/2^2 -1/4^2) =(1.09 xx 10^7 xx 12)/(4 xx 16)`

`1/lambda = 0.2057 xx 10^7`

`lambda = 4.861 xx 10^-7`

`lambda = 4861Å `