How do you calculate ionization energy using Rydberg constant?

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How do you calculate ionization energy using Rydberg constant?

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Answer 1 · 2016-08-06T20:52:23

$13.6 e V$ $13.6eV$ for first ionisation energy of hydrogen.

Explanation:

The Rydberg equation for absorption is

$\frac{1}{λ} = R (\frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}})$ $1/lambda = R(1/n_i^2 - 1/n_f^2)$

Where $λ$ $lambda$ is the wavelength of the absorbed photon, R is the Rydberg constant, $n_{i}$ $n_i$ denotes the energy level the electron started in and $n_{f}$ $n_f$ 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 $n_{f} = \infty$ $n_f = oo$ .

Assuming we ionise from the ground state, we set $n_{i} = 1$ $n_i = 1$

$\frac{1}{λ} = R$ $1/lambda = R$

$E = \frac{h c}{λ} \Rightarrow E = h c R$ $E = (hc)/lambda implies E = hcR$

$E = 6.626 \times 10^{- 34} \cdot 3 \times 10^{8} \cdot 1.097 \times 10^{7} = 2.182 \times 10^{- 18} J$ $E = 6.626xx10^(-34)*3xx10^8*1.097xx10^7 = 2.182xx10^(-18)J$

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

$1 e V = 1.6 \times 10^{- 19} J$ $1eV = 1.6xx10^(-19)J$ so to convert to eV we divide by $1.6 \times 10^{- 19}$ $1.6xx10^(-19)$

$\frac{2.182 \times 10^{- 18}}{1.6 \times 10^{- 19}} = 13.6 e V$ $(2.182xx10^(-18))/(1.6xx10^(-19)) = 13.6eV$

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How do you calculate ionization energy using Rydberg constant?

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