Question

If black body $A$ of surface area $S$ emits $32$ timesenergy per second than that emitted by another black body $B$ of surface area of $\frac{S}{2}$, then (${\lambda }_m$ is wavelength emitted with maximum energy proportion)

• A. $T_A=T_B$
• B. $T_B=2T_A$
• C. ${\lambda }_{m_B}=2{\lambda }_{m_A}$
• D. $e_B=2e_A$ ($e$ is emissivity)

Answer 1

The correct option is C ${\lambda }_{m_B}=2{\lambda }_{m_A}$

According to Stefan's law, the rate of energy transfer by a black body is given by,

$H=\sigma S{T}^4$

Given, $H_A=32H_B$

$\Rightarrow \sigma S{T}_A^4=32\sigma \frac{S}{2}{T}_B^4$

$\Rightarrow T_A=2T_B$

According to Wein's displacement law, for a black body,

$\lambda T=\text{constant}$,

$\Rightarrow {\lambda }_{m_A}{T}_A={\lambda }_{m_B}{T}_B$

$\Rightarrow 2{\lambda }_{m_A}={\lambda }_{m_B}$

Since both are black bodies,

$e_A=e_B=1$

Hence, $\left(C\right)$ is the correct answer.