Question

Two bodies A and B of equal surface area have thermal emissivities of $0.01$ and $0.81$ respectively. The two bodies are radiating energy at the same rate. Maximum energy is radiated from the two bodies A and B at wavelengths ${\lambda }_A$, and ${\lambda }_B$ respectively. Difference in these two wavelengths is 1 $\mu$. If the temperature of the body A is $5802K$, then value of ${\lambda }_B$ is :

• A. $\frac{3}{2}\mu m$
• B. $1\mu m$
• C. $2\mu m$
• D. $\frac{3}{4}\mu m$

Answer 1

Answer: (A)

$\frac{3}{2}\mu m$

We know that as per stephan's boltzman radiation law, $P\alpha \sigma A{T}^4$.
Since surface area of two bodies is same,

Therefore, ${\sigma }_A{T}_A^4={\sigma }_B{T}_B^4$. Hence, $T_A=3T_B$.

Now, as per Wein's displacement law, $\lambda T=constant=k$, $\lambda =\frac{k}{T}$.

${\lambda }_B-{\lambda }_A=k(\frac{1}{T_B}-\frac{1}{T_A})=k(\frac{1}{T_B}-\frac{1}{3T_B})=\frac{2k}{3T_B}=1\mu$

$T_B=\frac{T_A}{3}=1934K$

Putting in the above equation to calculate $k$ and the solving for ${\lambda }_B=\frac{k}{T_B}$, we get ${\lambda }_B=1.5\mu m$.