Question

An orange emission line is observed at 656.3 nm in the hydrogen emission spectrum. Determine the initial energy level of the electron.

Answer 1

Here, there is an emission line at λ= 656.3 nm (656.3 x 10^-9 m)

Color of the emitted photon is orange, therefore, final energy level is n = 2

We have,

ΔE=hc/ λ =[(6.62608 x 10^-34J s) x (2.99792 x 10⁸ ms^-1)] / (656.3 x 10^-9 m) = 3.0267 x 10^-19J

ΔE = [(1/n_final²) - (1/n_inital²)] RH

3.0267 x 10^-19J = [(1/2²)-(1/n_inital²)] x 2.18 x 10^-18J

0.1388 = [(1/4) - (1/n_inital²)]

(1/n_inital²) = (1/4) - 0.1388

(1/n_inital²) = 0.112

n_inital² = 8.9 9

Therefore, initial energy level of the electron = 3