Question

The temperature of a furnace is ${{2324}^{o}}C$ and the intensity is maximum in its spectrum nearly at nearly at $12000\overset{\text{o}}{\mathop{\text{A}}}\,$. If the intensity in the spectrum of star is maximum nearly at $4800\overset{\text{o}}{\mathop{\text{A}}}\,$, then the surface temperature of the star is:
(A). ${{8400}^{o}}C$
(B). ${{6219}^{o}}C$
(C). ${{7200}^{o}}C$
(D). ${{5900}^{o}}C$

Answer 1

Hint: A furnace being heated at a certain temperature and a star are considered as black bodies. Therefore, according to the Wien’s displacement law, the wavelength of maximum intensity for radiation at a particular temperature is inversely proportional to the temperature. Using this relation, we can calculate the temperature on the surface of a star.
Formulas used:
$\dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\dfrac{{{T}_{2}}}{{{T}_{1}}}$

Complete answer:
When a black- body is heated, it starts emitting radiation of different wavelengths at different temperatures.
According to the Wien’s displacement law, the wavelength maximum intensity of radiation emitted by a black body is inversely proportional to the temperature. This means for a particular temperature, as the wavelength increases, the intensity of the radiation decreases.
Therefore,
$\lambda \propto \dfrac{1}{T}$
Here, $\lambda $ is the wavelength of the radiation emitted
$T$ is the temperature
Therefore, from the above equation,
$\dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\dfrac{{{T}_{2}}}{{{T}_{1}}}$ - (1)
Given, let the temperature of a furnace equal to ${{2324}^{o}}C$ be ${{T}_{1}}$ and the temperature on the surface of a star be ${{T}_{2}}$. The wavelength of maximum intensity for furnace equal to $12000\overset{\text{o}}{\mathop{\text{A}}}\,$ be ${{\lambda }_{1}}$ and the wavelength of maximum intensity for spectrum of a star equal to $4800A$ be ${{\lambda }_{2}}$.
We substitute given values in eq (1) equation, to get,
$\begin{align}
  & \dfrac{12000\times {{10}^{-10}}}{4800\times {{10}^{-10}}}=\dfrac{{{T}_{2}}}{(2324+273)} \\
 & \Rightarrow {{T}_{2}}=\dfrac{10\times 2597}{4} \\
 & \Rightarrow {{T}_{2}}=6492.5K \\
 & \Rightarrow {{T}_{2}}=6492.5-273 \\
 & \Rightarrow {{T}_{2}}={{6219.5}^{o}}C \\
 & \therefore {{T}_{2}}\approx {{6219}^{o}}C \\
\end{align}$
The temperature of the surface of the star comes out to be ${{6220}^{o}}C$
Therefore, the temperature of the surface of the star is ${{6220}^{o}}C$.

Hence, the correct option is (B).

Note:
The star is often considered as a black body because the radiation given out is similar to a black body radiation. All those bodies which can absorb radiations along every wavelength are known as black bodies. For a black body to be in equilibrium, it must emit radiation at the same rate it absorbs it.