Question

The hot glowing surfaces of stars emit energy in the form ofelectromagnetic radiation. It is...

The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume that the emissivity e is equal to 1 for these surfaces.

Part A

Find the radius RRigel of the star Rigel, the bright blue star in the constellation Orion that radiates energy at a rate of 2.7×10^31 W and has a surface temperature of 11,000 K. Assume that the star is spherical.

Use σ=5.67×10^−8 W/m2⋅K4 for the Stefan-Boltzmann constant and express your answer numerically in meters to two significant figures.

Part B

Find the radius RProcyonB of the star Procyon B, which radiates energy at a rate of 2.1×10^23 W and has a surface temperature of 10,000 K. Assume that the star is spherical.

Use σ=5.67×10^−8 W/m2⋅K4 for the Stefan-Boltzmann constant and express your answer numerically in meters to two significant figures.

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Answer #1

Part(A):-Rate of Energy emitted by Star Rigel

$$ \begin{array}{l} =e \sigma T^{4} \times \text { Area } \\ =e \sigma T^{4} \times 4 \pi R_{\text {Rigel }}^{2} \end{array} $$

put the values given in the problem in above equation we get-

\(2.7 \times 10^{31}=1 \times 5.67 \times 10^{-8} \times(11000)^{4} \times 4 \times 3.14 \times R_{\text {Rigel }}^{2}\)

\(\Rightarrow R_{\text {Rigel }}^{2}=\frac{2.7 \times 10^{31}}{1 \times 5.67 \times 10^{-8} \times(11000)^{4} \times 4 \times 3.14}\)

\(\Rightarrow R_{\text {Rigel }}^{2}=2.589 \times 10^{21}\)

\(\Rightarrow R_{\text {Rigel }}=5.088 \times 10^{10} \mathrm{~m}\)

\(\Rightarrow R_{\text {Rigel }}=5.1 \times 10^{10} \mathrm{~m}\)

(Up to two significant figures)

\(\operatorname{Part}(B):-\)

Rate of Energy emitted by Star Procyon \(\mathrm{B}\)

$$ \begin{array}{l} =e \sigma T^{4} \times A r e a \\ =e \sigma T^{4} \times 4 \pi R_{\text {ProcyonB }}^{2} \end{array} $$

put the values given in the problem in above equation we get-

\(2.1 \times 10^{23}=1 \times 5.67 \times 10^{-8} \times(10000)^{4} \times 4 \times 3.14 \times R_{\text {ProcyonB }}^{2}\)

\(\Rightarrow R_{\text {ProcyonB }}^{2}=\frac{2.1 \times 10^{23}}{1 \times 5.67 \times 10^{-8} \times(10000)^{4} \times 4 \times 3.14}\)

\(\Rightarrow R_{\text {ProcyonB }}^{2}=2.9488 \times 10^{13}\)

\(\Rightarrow R_{\text {ProcyonB }}=5.43 \times 10^{6} \mathrm{~m}\)

\(\Rightarrow R_{\text {ProcyonB }}=5.4 \times 10^{6} \mathrm{~m} \quad\) (Up to two significant figures)

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Answer #2
Given :
.
Energy radiated (Q) = 2.7 * 1032 W
.
Temp (T) = 11000 K ;
.
Stefan-Boltzmann law states that the energy flux by radiationis proportional to the forth power of the temperature:
.
q = e ·s · T4
.
The total energy flux at a spherical surface of Radius R is :
.
Q =q·p·R² =e·s·T4 * p R²

Hence the radius is :
.
R = v( Q /(e·s·T^4·p) )
.
= v(2.7 * 1032 W / (1 * 5.67 *10-8 W/m²K4 *(1100K)4 · p) )
.
= 3.22 *1013 m
.
Similarly calculate for (b)
. Power ( P )= 2.1*10^23W
Temperature ( T ) = 10,000 K
emissivity ( e ) = 1
P = e s AT4
A = P / e sT4
But the area of the star ( A ) = 4 pr2
4 p r2 = P / es T4
= v [ P / ( es T4 ) * ( 4p ) ]
= -----
answered by: ym
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