Question

D. The Balmer Series 1. When Balmer found his famous series for hydrogen in 1886, he was limited experimentally to wavelengths in the visible and near ultraviolet regions from 250 nm to 700 nm, so all the lines in his series lie in that region. a) Based on the entries in Table 2 and the transitions on your energy level diagram, what common characteristic do the lines in the Balmer series have? B) What would be the longest possible wavelength for a line in the Balmer series? λ = ___________nm C) What would be the shortest possible wavelength that a line in the Balmer series could have? Hint: What is the largest possible value of ΔE to be associated with a line in the Balmer series? λ = ___________nm The Ionization Energy of Hydrogen 1. In the normal hydrogen atom the electron is in its lowest energy state, which is called the ground state of the atom. The maximum electronic energy that an electron in a hydrogen atom can have is 0 kJ/mole, at which point it would essentially be removed from the atom and it would become a H+ ion. How much energy in kilojoules per mole does it take to ionize a hydrogen atom? ______________ kJ/mole a) The ionization energy of hydrogen is often expressed in units other than kJ/mole. What would it be in joules per atom? in electron volts per atom? (1 ev = 1.602 x 10-19 J) ______________J/atom; _____________ ev/atom

Answer 1

A)

The common characteristics of lines in Balmer series are:

-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

B)

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.

C)

The Balmer series are lines that result from electrons falling from higher energy levels to the 2nd energy level. In B part 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. Its most probably that the wavelenght of light given off in such a case will be less than 250 nm.

D)

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).

E)

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