Quantum theory — Question 269

ANSWER

(a) A continuous curve with peak at a shorter wavelength and a higher peak radiance than the 3000 K curve. (b) $966\,\mathrm{nm}$ at $3000\,\mathrm K$; $644\,\mathrm{nm}$ at $4500\,\mathrm K$.

Worked solution

(a) Increasing temperature shifts the per-wavelength radiance peak toward shorter wavelength and raises its peak. Keep the low-wavelength rise and long-wavelength tail; do not draw a single spectral line.

(b) $\lambda_{\max}=b/T$. This gives $2.898\times10^{-3}/3000=966\,\mathrm{nm}$ and $2.898\times10^{-3}/4500=644\,\mathrm{nm}$.

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

Accept sensible significant figures and consistent rounding from stated constants.

Part a: Draw a smooth black-body curve peaking to the left of the supplied peak.

[1 mark]

Part a: Show the higher peak and label the 4500 K curve.

[1 mark]

Part b: Determine the first peak wavelength.

[1 mark]

Part b: Determine the second peak wavelength.

[1 mark]

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