WJEC Chemistry for AS Level Student Book: 2nd Edition (Draft)

 1.2 Basic ideas about atoms

As the frequency increases, the lines get closer together because the energy difference between the shells decreases. Each line in the Lyman series (ultraviolet region) is due to electrons returning to the first shell or n = 1 energy level, while the Balmer series (visible region) is due to electrons returning to the second shell or n = 2 energy level.

n =3 n =4 n =5 n = ∞

Paschen series

Balmer series

n =2

410 434 486

656

Study point The various series in the infrared region are caused by electrons returning to the energy levels n = 3 (Paschen), n = 4 (Brackett), n = 5 (Pfund) and n = 6 (Humphreys). You do not need to know the names of these series.

Wavelength/nm

n =1

Lyman series

Ionisation energy of the hydrogen atom The spectral lines become closer and closer together as the frequency of the radiation increases until they converge to a limit. The convergence limit corresponds to the point at which the energy of an electron is no longer quantised. At that point the nucleus has lost all influence over the electron; the atom has become ionised. For the Lyman series, n = 1, the convergence limit represents the ionisation of the hydrogen atom. Measuring the convergent frequency (difference from n = 1 to n = ∞ ) allows the ionisation energy to be calculated using Δ E = hf . The value of Δ E is multiplied by Avogadro’s constant to give the first ionisation energy for a mole of atoms.

Exam tip The IE of a hydrogen atom can be shown on its electron energy level diagram by drawing an arrow upwards from the n = 1 to the n = ∞ level. Key term The convergence limit is when the spectral lines become so close together they have a continuous band of radiation and separate lines cannot be distinguished.

Worked example The value of the wavelength at the start of the continuum in the hydrogen emission 29 DRAFT spectrum is 92nm. Calculate the first ionisation energy of hydrogen. (Assume that c = 3.00 × 10 8  ms −1 , h = 6.63 × 10 −34  Js and N A = 6.02 × 10 23  mol −1 ) Ionisation energy = N A Δ E ( N A = Avogadro’s constant) But Δ E = hf and f = c / λ ( h = Planck’s constant, c = speed of light) Therefore ionisation energy = N A hc / λ = 6.02 × 10 23 × 6.63 × 10 −34 × 3.00 × 10 8 (92 × 10 −9 ) = 1301498Jmol −1 = 1301kJmol −1 Knowledge check The value of the frequency at the start of the continuum in the sodium emission spectrum is 1.24 × 10 15  Hz. Calculate the first ionisation energy of sodium in kJ mol –1 . 14 don't forget to change 'nm' to 'm' by multiplying by 10 –9 Maths tip Wavelength has to be in metres . Don't forget to change 'nm' to 'm' by multiplying by 10 –9 .