Question

The temperature of a furnace is ${{2327}^{0}}C$ and the intensity in its spectrum is nearly at 12000A. If the intensity in the spectrum of a star is maximum nearly at 4800 A, then the surface temperature of the star is
A. ${{923}^{0}}C$
B. ${{1040}^{0}}C$
C. ${{6500}^{0}}C$
D. ${{6227}^{0}}C$

Answer 1

Hint: The relationship between temperature and maximum intensity that is lambda is given by Wien’s law and it is also called as Wien’s displacement law. Maximum wavelength states intensity emitted by light. Basically this law gives radiation that the body emits at a given temperature.

Complete step-by-step answer:
First let’s understand what the question wants to convey.
Question explanation- We have a furnace whose temperature is at 2327 degree Celsius. At this temperature its maximum intensity is 12000 ${{A}^{0}}$ . Now furnace intensity is at 48000 ${{A}^{0}}$ , then what will be its temperature when it has maximum 48000 ${{A}^{0}}$ intensity.

Aim- Find temperature at 48000 ${{A}^{0}}$ maximum intensity of furnace?
We know that,
${{\lambda }_{\max }}T=cons\tan t$
Let ${{T}_{1}}$ be the temperature of the furnace which is 2327 degrees when maximum intensity is 12000 ${{A}^{0}}$ .
${{\lambda }_{1}}$ is wavelength which is 12000 ${{A}^{0}}$ at 2327 degree temperature.
${{T}_{2}}$ be the temperature of the furnace when maximum intensity is 48000 ${{A}^{0}}$ .
${{\lambda }_{2}}$ is wavelength which is 48000 ${{A}^{0}}$ .
Notice that our temperature value is in degree, so convert it in Kelvin.

Conversion-
${{T}_{1}}$=2327+273=2600 kelvin
And automatically our ${{T}_{2}}$ will be in kelvin. We don’t need to convert ${{T}_{2}}$in kelvin.
So, ${{T}_{1}}$ be the temperature of the furnace which is 2600 degrees when maximum intensity is 12000 ${{A}^{0}}$ .
By Wien’s law,
$\begin{aligned} & {{\lambda }_{1}}{{T}_{1}}=cons\tan t \\\ & and \\\ & {{\lambda }_{2}}{{T}_{2}}=cons\tan t \\\ \end{aligned}$
Now equate both the equation we get,
${{\lambda }_{1}}{{T}_{1}}={{\lambda }_{2}}{{T}_{2}}$
Now put values

& 12000\times 2600=48000{{T}_{2}} \\\ & {{T}_{2}}=650kelvin(K) \\\ \end{aligned}$$ But our options are in degree so convert again into degree. $${{T}_{2}}=650+273={{923}^{0}}C$$ Temperature at 48000 $ {{A}^{0}} $ maximum intensity of furnace is $${{923}^{0}}C$$. Answer- (A) Note: This question is pretty straight forward, but while calculating you need to be careful because a lot of conversion is going on. Note that while calculating, you must have converted temperature in kelvin from degree. Then do calculate. Once you are done with calculation look for options. If options are in kelvin then keep answer as it is otherwise if option is in degree then convert it in degree by adding 273 in temperature value which was in degree.