Question

Three black bodies are radiating EM waves such that the higher intensity wavelengths are in the ratio $\lambda_{m1}:\lambda_{m2}:\lambda_{m3} = 1:(21)^{1/3}:(3)^{1/2}$. Which of the these is true for the temperatures?

• A. $T_1 > T_3 > T_2$
• B. $T_1 > T_2 > T_3$
• C. $T_3 > T_2 > T_1$
• D. $T_3 > T_1 > T_2$

Answer 1

Answer: (A)

$T_1 > T_3 > T_2$

Answer 2

For a black body, Wien’s displacement law gives

$\lambda_m T = b \quad \Rightarrow \quad T = \frac{b}{\lambda_m}$

Thus temperature is inversely proportional to $\lambda_m$. Given the ratios

$\lambda_{m1} : \lambda_{m2} : \lambda_{m3} = 1 : 21^{\frac{1}{3}} : 3^{\frac{1}{2}}$,

we approximate:

$21^{\frac{1}{3}} \approx 2.76, \quad 3^{\frac{1}{2}} \approx 1.73$.

This implies:

$\lambda_{m1} = 1,\quad \lambda_{m2} \approx 2.76,\quad \lambda_{m3} \approx 1.73$.

Then the temperatures are:

$T_1 \propto \frac{1}{1} = 1,\quad T_2 \propto \frac{1}{2.76} \approx 0.362,\quad T_3 \propto \frac{1}{1.73} \approx 0.578$.

Thus, the ordering is:

$T_1 > T_3 > T_2$.