College Physics
College Physics
10th Edition
ISBN: 9781285761954
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Chapter 27, Problem 6P

(a)

To determine

The energy of the photon.

(a)

Expert Solution
Check Mark

Answer to Problem 6P

The energy of the photon is 2.90×1019J/photon .

Explanation of Solution

The energy of the photon is,

EPeak=hcλmax

    • h is the Plank’s constant
    • λmax is the maximum wavelength
    • c is the speed of the light

Substitute 6.63×1034J.s for h , 3.00×108m/s for c and 685nm for λmax .

EPeak=(6.63×1034J.s)(3.00×108m/s)(685nm)(109m1nm)=2.90×1019J/photon

Conclusion:

Thus, the energy of the photon is 2.90×1019J/photon .

(b)

To determine

The surface temperature of the star.

(b)

Expert Solution
Check Mark

Answer to Problem 6P

The surface temperature of the star is 4.23×103K .

Explanation of Solution

According to Wein’s displacement law, the temperature is,

T=0.2898×102m.Kλmax

    • λmax is the maximum wavelength

Substitute 685nm for λmax .

T=0.2898×102m.K(685nm)(109m1nm)=4.23×103K

Conclusion:

Thus, the surface temperature of the star is 4.23×103K .

(c)

To determine

The rate at which energy is emitted from the star in the form of radiation.

(c)

Expert Solution
Check Mark

Answer to Problem 6P

The rate at which energy is emitted from the star in the form of radiation is 1.65×1026W

Explanation of Solution

The power radiated by an object is,

P=σAeT4=σ(4πr2)eT4

    • σ is the Stefan’s constant
    • A is the surface area
    • T is the temperature
    • e is the emissivity
    • r is the radius

Substitute 4.23×103K for T , 5.6696×108W/m2.K4 for σ , 1 for e and 8.50×108m for r .

P=(5.6696×108W/m2.K4)4π(8.50×108m)2(4.23×103K)4(1)=1.65×1026W

Conclusion:

Thus, the rate at which energy is emitted from the star in the form of radiation is 1.65×1026W .

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