Quantum theory — Question 277

ANSWER

(a) Absorption upward from -8 to -1 eV; emission downward from -1 to -5 eV and from -5 to -8 eV. (b) $310\,\mathrm{nm}$ and $413\,\mathrm{nm}$; the lower-energy 3.00 eV transition gives the longer wavelength.

Worked solution

(a) Absorption increases atomic energy; emission decreases it. The two downward arrows must connect the actual intermediate state.

(b) The gaps are $4.00$ and $3.00\,\mathrm{eV}$. Hence $\lambda=1240/4.00=310\,\mathrm{nm}$ and $1240/3.00=413\,\mathrm{nm}$.

Equivalent physically justified methods, alternative valid sketches and consistent units accepted.

Accept sensible significant figures and consistent rounding from stated constants.

Part a: Draw the correct upward absorption arrow.

[1 mark]

Part a: Draw and identify both downward cascade arrows.

[1 mark]

Part b: Calculate the 4.00 eV transition wavelength.

[1 mark]

Part b: Calculate the 3.00 eV transition wavelength and identify it as longer.

[1 mark]

Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.