# Balmer series for atomic hydrogen lie

• xregina12
In summary, the conversation discusses finding the Rydberg constant and minimum ionization energy for atomic hydrogen using the equations 1/lambda = R (1/4 - 1/n^2) and E=hc/lambda=hcR(1/4-1/n^2). The ionization limit corresponds to n → ∞ for the final state of the electron. The value of n for the Balmer series starts at 3 because it is the name given to spectral line emissions of the hydrogen atom. A photon is absorbed when the electron goes from n=1 to n=2.

## Homework Statement

The first few lines of the Balmer series for atomic hydrogen lie at the wavelengths λ = 656.46, 486.27, 434.17, 410.29nm, ... Find a value for RH, the Rydberg constant for hydrogen. The ionization energy I is the minimum energy required to remove the electron. Find it from the data. How is I related to RH? (Hint: the ionization limit corresponds to n → ∞ for the final state of the electron).

I used the equation: 1/lambda = R (1/4 - 1/n^2) where R is Rydberg constant and n=3, 4, 5 respectively and got that R=1.097 x 10 ^7. However, I am not sure what to do to find the minimum ionization energy. The only equation I thought I can use is E=hc/lambda=hcR(1/4-1/n^2). For Also I know that since it's the minimum energy required to remove the electron, I should use the longest wavelength given to figure out the ionization energy. Can someone please explain to me the second part of the problem to me? Also, why does n for the Balmer series start at 3? Please help me. Thank you!

xregina12 said:

## Homework Statement

However, I am not sure what to do to find the minimum ionization energy. The only equation I thought I can use is E=hc/lambda=hcR(1/4-1/n^2).
Look at the hint.