Atomic spectra and the energy of light: significance of the frequency

Overview

Twelve recall questions on atomic spectra and the energy of light: significance of the frequency.

What you need to know

Answer each one before you reveal it.

Questions from this subtopic

Each one with the model answer underneath.

1. 1

   What does the convergence limit of the Lyman series represent physically?

   Model answer

   It is the transition between n = ∞ and n = 1. Read in the direction of absorption, it is the energy needed to take the electron from the ground state right out of the atom, so it corresponds exactly to the ionisation of a hydrogen atom.

2. 2

   Why is the convergence limit of the Lyman series used to find the ionisation energy of hydrogen, rather than the limit of the Balmer series?

   Model answer

   Ionisation energy is defined from the ground state, which for hydrogen is n = 1, and only the Lyman series has n = 1 as its lower level. The Balmer limit measures the energy to remove an electron already excited to n = 2, which is smaller and is not the ionisation energy.

3. 3

   The convergence limit of the Lyman series occurs at a frequency of 3.28 × 10¹⁵ Hz. Calculate the ionisation energy of a single hydrogen atom.

   Model answer

   E = hf = 6.63 × 10⁻³⁴ J s × 3.28 × 10¹⁵ s⁻¹ = 2.17 × 10⁻¹⁸ J. This is the energy needed to remove the electron from one hydrogen atom in its ground state.

4. 4

   Using a convergence limit frequency of 3.28 × 10¹⁵ Hz, calculate the ionisation energy of hydrogen in kJ mol⁻¹.

   Model answer

   The energy per atom is E = hf = 6.63 × 10⁻³⁴ × 3.28 × 10¹⁵ = 2.17 × 10⁻¹⁸ J. Multiplying by the Avogadro constant gives 2.17 × 10⁻¹⁸ J × 6.02 × 10²³ mol⁻¹ = 1.31 × 10⁶ J mol⁻¹, which is 1.31 × 10³ kJ mol⁻¹, three significant figures being justified by the data.

5. 5

   What wavelength corresponds to a convergence limit frequency of 3.28 × 10¹⁵ Hz, and in which region of the spectrum does it lie?

   Model answer

   λ = c/f = (3.00 × 10⁸ m s⁻¹) ÷ (3.28 × 10¹⁵ s⁻¹) = 9.15 × 10⁻⁸ m, that is 91.5 nm. This is well below the shortest visible wavelength, so it lies in the ultraviolet.

6. 6

   Why must the ionisation energy of hydrogen obtained from a spectrum be multiplied by the Avogadro constant before it can be compared with a tabulated value?

   Model answer

   The spectroscopic calculation gives the energy for a single atom, in joules, whereas a tabulated ionisation energy refers to one mole of gaseous atoms. Multiplying by 6.02 × 10²³ mol⁻¹ converts one to the other, and dividing by 1000 then gives kJ mol⁻¹.

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