Question

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

• A. $7219.5^{\circ }C$
• B. $6219.5^{\circ }C$
• C. $6319.5^{\circ }C$
• D. None of these

Answer 1

Answer: (B)

From Wien's displacement law
${\lambda }_{m}T=$ constant
Given, $T_1=2324^{\circ }C=2597K$,
${\lambda }_{m_1}=12000\mathring{A}$,
${\lambda }_{m_2}=4800\mathring{A}$
$\therefore {\lambda }_{m_1}{T}_1={\lambda }_{m_2}{T}_2$
$\Rightarrow T_2=\frac{{\lambda }_{m_1}}{{\lambda }_{m_2}}{T}_1$
$\Rightarrow T_2=\frac{12000}{4800}\times 2597$
$=6492.5K$
$\Rightarrow T_2=(6492.5-273{)}^{\circ }C$
$=6219.5^{\circ }C$