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Balmer Series and Ionization Energy of Hydrogen

Submitted by needhelp on Fri, 01/04/2008 - 21:26

I have a worksheet that deals with Balmer Series. I did most of it but there were a few that confused me a little:

1. What common characteristic do the lines in the Balmer Series have? (I think it has to do with their energies, but I'm not sure)

2. What would be the longest possible wavelength for a line in the Balmer series? (It would be the one with the lowest ENERGY or WAVELENGTH?

3. Fundamentally, why would any line in the hydrogen spectrum between 250 nm and 700 nm belong to the Balmer series?

4. The maximum electronic energy that a hydrogen atom (normally ground state) can have is O kj/mole, at which point the electron would essentially be removed from the atom and it would become a H+ ion. How much energy in kj/mole does it take to ionize an H atom.

5. The ionization energy of hydrogen is often expressed in units other than kj/mole. What would it be joules per atom? in electron volts per atom? (1 ev=1.602 X 10-19)

Not positive, but this is what I would think they might be looking for.
1) The Balmer series are lines in the visible light range of the electromagnetic spectrum.  They result from electron transitions from higher energy levels to the 2nd energy level in the hydrogen atom
2) The electron transition which would produce the smallest energy in the Balmer series would be from the 3rd energy level to the 2nd energy level.  This transition results in a photon with a wavelenght of 656 nm.
3) The Balmer series are lines that result from electrons falling from higher energy levels to the 2nd energy level.  In 2 above we said the longest wavelength would be 656 nm (less than 700 nm).  The greatest amount of energy would result when an electron falls from the outermost energy level  down to the 2nd energy level.  I'm pretty sure that the wavelenght of light given off in such a case will be less than 250 nm.
4)  1312 kJ/mol  (Chemistry by McMurray & Fay)  This could also be calculated by determining the energy of the electron in the 1st energy level of a hydrogen atom using Bohr's equation:
E= -Ze2/2r (where Z is the atomic #, e is the charge on the electron, and r is the radius of the atom.
5)   Convert kJ/mole to J/atom:
                 1312 kJ/mol  x  1mol/6.02E23   x 1E3J/kJ
      Then convert J/atom to ev/atom:
                      J/atom   x    1 ev/1.602E-19

just remember that the visible spectrum of light from hydrogen displays four wave length; 410 nm , 434 nm, 486 nm and 656 nm, that reflect the emissions of photons by electrons in exited states transitioning to the quantum level described by principle quamtum number ( n=2).

i hope that helps for third answer.