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, $T_{1}=2324^{\circ} C =2597\, K$,
$\lambda_{m_{1}}=12000 \,\mathring{A}$,
$\lambda_{m_{2}}=4800 \,\mathring{A}$
$\therefore \lambda_{m_{1}} T_{1}=\lambda_{m_{2}} T_{2}$
$\Rightarrow T_{2}=\frac{\lambda_{m_{1}}}{\lambda_{m_{2}}} T_{1}$
$\Rightarrow T_{2}=\frac{12000}{4800} \times 2597$
$=6492.5 \,K$
$\Rightarrow T_{2}=(6492.5-273)^{\circ} C$
$=6219.5^{\circ} C$