Answer :
The blackbody radiation energy emitted by star a is 1/16 factor of star b energy.
We need to know about black body radiation to solve this problem. The energy radiated by a black body object is proportional to the area and the fourth power of temperature. It can be determined as
P = A . e . σ . T⁴
where P is power, A is surface area, e is emissivity, σ is Stefan Boltzmann's constant ( 5.67 x 10¯⁸ W/m²K⁴) and T is temperature.
From the question above, we know that
Tb = 2 Ta
By using the black body radiation formula, the ratio of temperature is
Ta⁴ / Tb⁴ = Pa / Pb
Ta⁴ / Tb⁴ = Ea / Eb
Hence,
Ta⁴ / Tb⁴ = Ea / Eb
Ta⁴ / (2Ta)⁴ = Ea / Eb
Ta⁴ / 16Ta⁴ = Ea / Eb
1/16 . Eb = Ea
Thus, the energy of star a will have 1/16 factor of star b energy
Find more on black body radiation at: https://brainly.com/question/18273651
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