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A star which appears blue will be
A) much hotter than the Sun
B) colder than the Sun
C) as hot as the Sun
D) at $ - {273^ \circ }C$

Last updated date: 20th Jun 2024
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Hint: Energy is inversely proportional to the wavelength and is directly proportional to the temperature of the body. So just check the relation and compare temperature. The light colour with less wavelength will have more temperature.

Complete step by step answer:
Explanation of the symbols used:
 $h$ = Planck’s constant
 $c$ = speed of light
 ${E_{star}}$ = energy per unit area (flux) radiated by the blue star
 ${E_{sun}}$ = energy per unit area (flux) radiated by the Sun
 $\sigma $ = Stefan-Boltzmann constant
 $\varepsilon $ = emissivity of the body
 ${T_{star}}$ = temperature of the blue star
 ${T_{sun}}$ = temperature of the Sun

We know that energy $E$ is inversely proportional to the wavelength $\lambda $ according to the relation
  $E = \dfrac{{hc}}{\lambda }$.
Hence as we know the wavelength of blue light is less than the wavelength of yellow or red light (colour of the Sun). The energy of the blue star is therefore more than the energy of the Sun (since it is inversely proportional).

 ${E_{star}} > {E_{sun}}$
Hence according to Stefan-Boltzmann Law we know that $E = \sigma \varepsilon {T^4}$
$\sigma \varepsilon T_{star}^4 > \sigma \varepsilon T_{sun}^4$
${T_{star}} > T_{sun}^{}$
Therefore, the temperature of the star will be greater than the temperature of the Sun, i.e., the star which appears blue will be much hotter than the Sun.
Hence the correct answer is option A.

Note: The comparison can be made by Wien’s displacement law as well which directly relates the wavelength of emitted radiation and temperature of a body as
 $\lambda T$ = constant
Which means since blue light has less wavelength as compared to Yellow or Orange light, it will have more temperature, i.e., it will be hotter. This method can be used since it is a shorter method but our solution that we have discussed is more fundamental in nature and hence generally recommended.