How do you calculate ionization energy using Rydberg constant?

1 Answer
Aug 6, 2016

13.6eV for first ionisation energy of hydrogen.

Explanation:

The Rydberg equation for absorption is

1/lambda = R(1/n_i^2 - 1/n_f^2)

Where lambda is the wavelength of the absorbed photon, R is the Rydberg constant, n_i denotes the energy level the electron started in and n_f the energy level it ends up in.

We are calculating ionisation energy so the electron goes to infinity with respect to the atom, ie it leaves the atom. Hence we set n_f = oo.

Assuming we ionise from the ground state, we set n_i = 1

1/lambda = R

E = (hc)/lambda implies E = hcR

E = 6.626xx10^(-34)*3xx10^8*1.097xx10^7 = 2.182xx10^(-18)J

When we deal with such small energies, it is often helpful to work in electron volts.

1eV = 1.6xx10^(-19)J so to convert to eV we divide by 1.6xx10^(-19)

(2.182xx10^(-18))/(1.6xx10^(-19)) = 13.6eV