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Wave/particle duality

Q4. The diagram shows some of the electron energy levels of an atom.

An incident electron of kinetic energy 4.1 × 10–18 J and speed 3.0 × 106 m s–1 collides with the atom represented in the diagram and excites an electron in the atom from level B to level D.

(a) For the incident electron, calculate

(i) the kinetic energy in eV,

Ek. = 4.1 × 10–18/1.6 x 10-19 = 25.6 eV

Ek = 26 eV (values are given to 2sf only - so your answer should be 2sf too)

(ii) the de Broglie wavelength.

λ dB= h/p = h/mv

λ dB= 6.63 x 10-34/ (9.11 x 10-31 x 3.0 × 106)

λ dB= 2.42 × 10–10 m

λ dB= 2.4 × 10–10 m

(4 marks)

(b) When the excited electron returns directly from level D to level B it emits a photon. Calculate the wavelength of this photon.

ΔE = (-0.21 - (-0.90)) x 10-18 = 0.69 x 10-18J

ΔE = hf = hc/λ

λ = hc/ΔE

λ = (6.63 x 10-34 x 3.0 x 108)/0.69 x 10-18

λ = 2.9 x 10-7 m

(3 marks)

(Total 7 marks)