The emission spectrum of atomic hydrogen consists of a number of discrete wavelengths.

Question 1

(a) The emission spectrum of atomic hydrogen consists of a number of discrete wavelengths.

Explain how this observation leads to an understanding that there are discrete electron

energy levels in atoms. [2]

(b) Some electron energy levels in atomic hydrogen are illustrated in Fig. 7.1.

The longest wavelength produced as a result of electron transitions between two of the

energy levels shown in Fig. 7.1 is 4.0 × 10^-6 m.

(i) On Fig. 7.1,

1. draw, and mark with the letter L, the transition giving rise to the wavelength of

4.0 × 10^-6 m, [1]

2. draw, and mark with the letter S, the transition giving rise to the shortest

wavelength. [1]

(ii) Calculate the wavelength for the transition you have shown in (i) part 2. [3]

(c) Photon energies in the visible spectrum vary between approximately 3.66 eV and

1.83 eV.

Determine the energies, in eV, of photons in the visible spectrum that are produced by

transitions between the energy levels shown in Fig. 7.1. [2]

Reference: Past Exam Paper – June 2013 Paper 42 Q7

Solution:

(a)

Each wavelength is associated with a discrete change in energy.

The discrete energy change / difference implies discrete levels.

(b)

(i)

1.

{For emission, the transition is from the higher energy level to lower energy level.}

ΔE = hc / λ   (longest wavelength corresponds to smallest energy change)

Arrow from – 0.54eV to – 0.85eV, labeled L           

2.

Arrow from – 0.54eV to – 3.4eV, labeled S

(ii)

ΔE = hc / λ 

(3.4 – 0.54) × 1.6×10^-19 = (6.63×10^-34 × 3.0×10⁸) / λ 

λ = 4.35×10^-7 m

(c)

-1.50 to -3.4 = 1.9 eV

-0.85 to -3.4 = 2.55 eV (allow 2.6eV)

-0.54 to -3.4 = 2.86 eV (allow 2.9eV)