Question

Two spherical stars A and B emit blackbody radiation. The radius of A is 400 times that of B and A emits ${{10}^{4}}$ times the power emitted from B. The ratio $\left( \dfrac{\lambda A}{\lambda B} \right)$ of their wavelengths $\lambda A$ and $\lambda B$ at which the peaks occur in their respective radiation curves is

Answer 1

We have two stars that emit black body radiation. We are given the relation between power and radius of these two stars. From Wien’s law we have a relation between wavelength and absolute temperature and from Stefan’s law we have an equation for power. By using these equations we will get the required solution.

Formula used:
$\lambda T=\text{constant}$
$P=\sigma eA{{T}^{4}}$

Complete step-by-step answer:
In the question we are given two spherical stars A and B which emits black body radiation.
It is said that the radius of the star A is 400times of the star B, i.e. if ‘${{R}_{A}}$’ is the radius of the star A and ‘${{R}_{B}}$’ is the radius of the star B, then we have
${{R}_{A}}=400{{R}_{B}}$
It is also said that the power emitted by A is ${{10}^{4}}$ times the power emitted by B, i.e. if ‘${{P}_{A}}$’ and ‘${{P}_{B}}$’ are the powers emitted by the stars A and B respectively, then we have
${{P}_{A}}={{10}^{4}}{{P}_{B}}$
From Wien’s law we know that,
$\lambda T=\text{constant}$, were ‘$\lambda$’ is the wavelength and ‘T’ is the absolute temperature.
Therefore in this case for star A and B, according to Wien’s law we can write
${{\lambda }_{A}}{{T}_{A}}={{\lambda }_{B}}{{T}_{B}}$
From this equation we get,
$\Rightarrow \dfrac{{{T}_{A}}}{{{T}_{B}}}=\dfrac{{{\lambda }_{B}}}{{{\lambda }_{A}}}$
Now from Stefan’s law we have,
$P=\sigma eA{{T}^{4}}$ , were ‘P’ is the power, ‘$\sigma$’ is Stefan’s constant, ‘e’ is emissivity, ‘A’ is area and ‘T’ is absolute temperature.
Therefore the power of the stars A and B according to Stefan’s law can be written as,
$\Rightarrow {{P}_{A}}=\sigma e{{A}_{A}}{{T}_{A}}^{4}$
$\Rightarrow {{P}_{B}}=\sigma e{{A}_{B}}{{T}_{B}}^{4}$
We know that for a black body, the emissivity e is 1. Therefore we get,
$\Rightarrow {{P}_{A}}=\sigma {{A}_{A}}{{T}_{A}}^{4}$
$\Rightarrow {{P}_{B}}=\sigma {{A}_{B}}{{T}_{B}}^{4}$
From the above two equations, by taking the ratio of ‘${{P}_{A}}$’ and ‘${{P}_{B}}$’, we get
$\Rightarrow \dfrac{{{P}_{A}}}{{{P}_{B}}}=\dfrac{\sigma {{A}_{A}}{{T}_{A}}^{4}}{\sigma {{A}_{B}}{{T}_{B}}^{4}}$
$\Rightarrow \dfrac{{{P}_{A}}}{{{P}_{B}}}=\dfrac{{{A}_{A}}{{T}_{A}}^{4}}{{{A}_{B}}{{T}_{B}}^{4}}$
From earlier application of Wien’s law we have,
$\dfrac{{{T}_{A}}}{{{T}_{B}}}=\dfrac{{{\lambda }_{B}}}{{{\lambda }_{A}}}$
By applying this here, we get
$\Rightarrow \dfrac{{{P}_{A}}}{{{P}_{B}}}=\dfrac{{{A}_{A}}}{{{A}_{B}}}{{\left( \dfrac{{{\lambda }_{B}}}{{{\lambda }_{A}}} \right)}^{4}}$
In the question we are asked to find the ratio of the wavelengths of the two stars $\left( \dfrac{{{\lambda }_{A}}}{{{\lambda }_{B}}} \right)$ . From the above equation we get this ratio as,
$\Rightarrow \dfrac{{{\lambda }_{A}}}{{{\lambda }_{B}}}={{\left( \dfrac{{{P}_{B}}{{A}_{A}}}{{{P}_{A}}{{A}_{B}}} \right)}^{{}^{1}/{}_{4}}}$
We know that the area of the stars A and B will be,
${{A}_{A}}=\pi {{R}_{A}}^{2}$
${{A}_{B}}=\pi {{R}_{B}}^{2}$
Therefore we get the ratio as,
$\Rightarrow \dfrac{{{\lambda }_{A}}}{{{\lambda }_{B}}}={{\left( \dfrac{{{P}_{B}}}{{{P}_{A}}}\dfrac{\pi {{R}_{A}}^{2}}{\pi {{R}_{B}}^{2}} \right)}^{{}^{1}/{}_{4}}}$
$\Rightarrow \dfrac{{{\lambda }_{A}}}{{{\lambda }_{B}}}={{\left( \dfrac{{{P}_{B}}}{{{P}_{A}}}\dfrac{{{R}_{A}}^{2}}{{{R}_{B}}^{2}} \right)}^{{}^{1}/{}_{4}}}$
From the question we know that,
${{R}_{A}}=400{{R}_{B}}$
${{P}_{A}}={{10}^{4}}{{P}_{B}}$
Therefore the ratio becomes,
$\Rightarrow \dfrac{{{\lambda }_{A}}}{{{\lambda }_{B}}}={{\left( \dfrac{{{P}_{B}}}{{{10}^{4}}{{P}_{B}}}\dfrac{{{\left( 400 \right)}^{2}}{{R}_{B}}^{2}}{{{R}_{B}}^{2}} \right)}^{{}^{1}/{}_{4}}}$
$\Rightarrow \dfrac{{{\lambda }_{A}}}{{{\lambda }_{B}}}={{\left( \dfrac{1}{{{10}^{4}}}\dfrac{{{\left( 400 \right)}^{2}}}{1} \right)}^{{}^{1}/{}_{4}}}$
$\Rightarrow \dfrac{{{\lambda }_{A}}}{{{\lambda }_{B}}}={{\left( \dfrac{{{\left( 400 \right)}^{2}}}{{{10}^{4}}} \right)}^{{}^{1}/{}_{4}}}$
$\Rightarrow \dfrac{{{\lambda }_{A}}}{{{\lambda }_{B}}}={{\left( \dfrac{16\times {{10}^{4}}}{{{10}^{4}}} \right)}^{{}^{1}/{}_{4}}}$
$\Rightarrow \dfrac{{{\lambda }_{A}}}{{{\lambda }_{B}}}={{\left( 16 \right)}^{{}^{1}/{}_{4}}}$
$\Rightarrow \dfrac{{{\lambda }_{A}}}{{{\lambda }_{B}}}=2$
Therefore the ratio of the wavelengths of the stars A and B is 2.

Note: A black body is a body that absorbs all the energies radiated on the surface of that body and re - -emits that radiation. This radiation that is re – emitted by the black body is known as black body radiation.
Wien’s law or Wien’s displacement law gives us the relation between the temperature and wavelength of radiation by a black body. The law simply states that the wavelength at the peak changes with temperature.
Stefan’s law or Stefan – Boltzmann law gives us the power radiated from a black body in terms of its absolute temperature. The law states that the total radiant heat power emitted from the surface of a black body is directly proportional to the fourth power of its temperature.