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Physics Question on thermal properties of matter

Ratio of radiant energy emitted per unit area per second by two stars is 16 : 81. The ratio of wavelengths at which intensity of radiant energy is maximum is given by?

A

9:04

B

4:09

C

2:03

D

3:02

Answer

3:02

Explanation

Solution

As , EE (energy radiated per unit area per second) ∝T4\propto T^4
∴  E1E2=T14T24      \therefore \, \, \frac{E_{1}}{E_{2} } = \frac{T_{1^{4}}}{T_{2^{4}}} \, \, \, \, \, \, ...(i)
Also, according to Wein's displacement law,
T1T2=(λm)2(λm)1    \frac{T_{1}}{T_{2} } = \frac{\left(\lambda_{m}\right)_{2}}{\left(\lambda_{m}\right)_{1}} \, \, \, \, ...(ii)
From eqn (i) and (ii),
E1E2=[(λm)2(λm)1]4\frac{E_{1}}{E_{2} } = \left[\frac{\left(\lambda _{m}\right)_{2}}{\left(\lambda _{m}\right)_{1}}\right]^{4}
Here , E1E2=1681\frac{E_{1}}{E_{2}} = \frac{16}{81}
So, (1681)14=(λm)2(λm)1or(λm)1(λm)2=32\left(\frac{16}{81}\right)^{1 4 } = \frac{\left(\lambda _{m}\right)_{2}}{\left(\lambda _{m}\right)_{1}} or \frac{\left(\lambda _{m}\right)_{1}}{\left(\lambda _{m}\right)_{2}} = \frac{3}{2}