Question

The temperature of a furnace is ${2324}^{\circ }C$ and the intensity is maximum in its radiation spectrum nearly at $12000$ $A^{\circ }$. If the intensity in the spectrum of a star is maximum nearly at $4800$ $A^{\circ }$, then the surface temperature of the star is :

• A. $8400K$
• B. $7200K$
• C. $6492.5K$
• D. ${5900}^{\circ }C$

Answer 1

Answer: (C)

$6492.5K$

${\lambda }_m\times T=constant$.
${\lambda }_{m_1}{T}_1={\lambda }_{m_2}{T}_2$
Given ${\lambda }_{m_1}=12000A^o$, $T_1={2324}^{o}C=2597K$, ${\lambda }_{m_2}=4800A^o$
$T_2=?$
${\lambda }_{m_1}{T}_1={\lambda }_{m_2}{T}_2{T}_2=\frac{{\lambda }_{m_1}{T}_1}{{\lambda }_{m_2}}=\frac{12000\times 2597}{4800}=6492.5K$
option (C) is the correct answer.