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Electron Configurations · Paper 2 — Free Chemistry SL Practice Questions

1ChallengeSAQ-LElectron arrangement and atomic orbitals7 marksPaper 2~11 min

Data

c = 3.00 \times 10^{8}\,\text{m s}^{-1}

;

h = 6.63 \times 10^{-34}\,\text{J s}

;

N_A = 6.02 \times 10^{23}\,\text{mol}^{-1}

;

R_H = 1.097 \times 10^{7}\,\text{m}^{-1}

. Formulae:

\dfrac{1}{\lambda} = R_H\!\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)

;

E = h\nu

;

c = \lambda\nu

The emission spectrum of atomic hydrogen contains a series of lines in the ultraviolet region known as the Lyman series. These lines correspond to electron transitions from higher energy levels (

n > 1

) to the ground state (

n = 1

(a)

Calculate the ionization energy of hydrogen in

\text{kJ mol}^{-1}

using the Rydberg equation. [3 marks]

(b)

Explain why the lines in the Lyman series converge at shorter wavelengths. [1 mark]

(c)

Calculate the wavelength, in nm, of the spectral line corresponding to the transition

n = 3 \to n = 1

. [2 marks]

(d)

Evaluate how the observation of discrete spectral lines and a convergence limit in the Lyman series supports the quantum mechanical model of the atom over the classical (continuous) model. [1 mark]

Solutions

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