Question

Two spherical bodies $A$ and $B$ with a radius of $12\text{cm}$ and $18\text{cm}$ respectively are at temperatures $T_A$ and $T_B$. The maximum intensity in the emission spectrum of $A$ is at $1000\text{nm}$ and in that of $B$ is at $2000\text{nm}$. Considering both bodies to be black bodies, Find the ratio of the rate of total energy radiated by $A$ to that of $B$ ?

• A. $7.1$
• B. $4.2$
• C. $3.6$
• D. $2.4$

Answer 1

The correct option is A $7.1$

Stefan-Boltzmann law gives the energy radiated per unit time by a spherical black body of area $A(=4\pi R^2)$ and temperature $T$ as
$E=\sigma A{T}^4=4\pi \sigma R^2{T}^4.....\left(1\right)$
Wien's displacement law, states that ,
${\lambda }_{m}T=$ Constant $=C.......\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$ By eliminating ${\lambda }_m$ we get,
$E=4\pi \sigma C^4\left(\frac{R^2}{{\lambda }_m^4}\right)$
$\Rightarrow E\propto \left(\frac{R^2}{{\lambda }_m^4}\right)$
From the data given in the question ,
$\frac{E_A}{E_B}={\left(\frac{R_A}{R_B}\right)}^2{\left(\frac{{\lambda }_B}{{\lambda }_A}\right)}^4={\left(\frac{12}{18}\right)}^2{\left(\frac{2000}{1000}\right)}^4=7.1$
Hence, option $\left(a\right)$ is the correct answer.