Question

A black body at temperature T is radiating energy at a rate $E_1$ through waves of wavelength range $\lambda_1$ to $\lambda_1 + \Delta\lambda$ and $E_2$ through wavelength range $\lambda_2$ to $\lambda_2 + \Delta\lambda$. If $E_1 = E_2$ [b: Wein's constant and $\lambda_2 > \lambda_1$, $\Delta\lambda<< \lambda_1$]

• A. $\frac{b}{\lambda_1}<T<\frac{b}{\lambda_2}$
• B. $\frac{b}{\lambda_1}>T>\frac{b}{\lambda_2}$
• C. $\frac{b}{\lambda_1}=T=\frac{b}{\lambda_2}$
• D. None of these

Answer 1

Answer: (B)

$\frac{b}{\lambda_1}>T>\frac{b}{\lambda_2}$

Answer 2

The energy radiated by a black body in a specific wavelength range is approximated by $B(\lambda, T) \Delta\lambda$. Given $E_1 = E_2$ and $\Delta\lambda << \lambda_1$, we have $B(\lambda_1, T) \approx B(\lambda_2, T)$. The spectral radiance curve $B(\lambda, T)$ peaks at $\lambda_{max} = b/T$. Since $\lambda_1 < \lambda_2$ and $B(\lambda_1, T) \approx B(\lambda_2, T)$, $\lambda_1$ must be to the left of the peak and $\lambda_2$ to the right. Thus, $\lambda_1 < \lambda_{max} < \lambda_2$, which means $\lambda_1 < b/T < \lambda_2$. Rearranging this inequality gives $\frac{b}{\lambda_2} < T < \frac{b}{\lambda_1}$, or equivalently, $\frac{b}{\lambda_1} > T > \frac{b}{\lambda_2}$.