Suppose a star with radius 8.50 ×10^8 m has a peak wavelength of 685 nm in the spectrum of its e

Suppose a star with radius 8.50 ×10^8 m has a peak...

Suppose a star with radius $8.50 \times 10^{8} \mathrm{m}$ has a peak wavelength of 685 $\mathrm{nm}$ in the spectrum of its emitted radiation. (a) Find the energy of a photon with this wavelength. (b) What is the surface temperature of the star? (c) At what rate is energy emitted from the star in the form of radiation? Assume the star is a blackbody $(e=1)$ . (d) Using the answer to part (a), estimate the rate at which photons leave the surface of the star.

Key Concepts

Photon Emission Rate

Once the energy of individual photons is known, the concept of photon emission rate involves calculating the number of photons emitted per unit time by dividing the total radiated power by the energy per photon. This concept bridges the quantum and thermodynamic descriptions of radiative processes in astrophysical objects.

Stefan-Boltzmann Law

The Stefan-Boltzmann Law governs the total power radiated per unit area from a blackbody in terms of its temperature. Expressed as P = ?A T^4, where ? is the Stefan-Boltzmann constant, this law is essential for understanding how energy is emitted from stars and other astronomical objects as electromagnetic radiation.

Photon Energy Calculation

This concept is based on Planck's relation, which explains that the energy of a photon is inversely proportional to its wavelength. The formula E = hc/?, where h is Planck's constant and c is the speed of light, is fundamental in quantum mechanics and reveals the quantized nature of electromagnetic radiation.

Wien's Displacement Law

Wien's Displacement Law describes the relationship between the temperature of a blackbody and the wavelength at which it emits radiation most intensely. It states that the peak wavelength is inversely proportional to the temperature (?_max * T = constant), and is crucial for determining the temperature of stars and other radiating bodies.

Transcript

00:01 Okay, so in this problem, we're told that a star has a radius of 8 .5 times 10 of the 8.

00:16 And we want, what else are we told? and the peak wavelength is 685.

00:22 So lambda is 685 nanometers.

00:30 And we want to get the energy of a photon with that wavelength.

00:35 All right, we can just use e is hf.

00:39 And then f is c over lambda.

00:42 So he is hc over lambda, h is plunks constant, c is a speed of light, and lambda is this lambda right here.

00:49 So i'm going to go ahead and plug that into a calculator, 6 .626 times 10 to the minus 34 times 2 .998 times 10 to the 8th, all divided by 685 times 10 to the minus 9.

01:08 And so with that, i get, oh, is this right? c, 2 .99, 3 .78.

01:19 Okay, 5.

01:21 I get 2 .896 many other digits times 10 of the minus 19 joules.

01:29 So i will go ahead and write that out.

01:32 So 2 .90.

01:34 Let's see, 3 sigphigs, 2 .90 times 10 to the minus 19 tools.

01:47 Just double check that digit.

01:57 And we want to get the surface temperature of the star so we can figure that out using wines displacement law so for that what we want to do is say that lambda times t is equal to 0 .2898 times 10 to the minus 2 kelvin meters and we want to take both sides and divide it by lambda to get the temperature so i'll just go ahead and erase that so then we can get that the temperature is equal to this.

02:33 And i'm going to go ahead and plug that into a calculator.

02:39 0 .2898 times 10 to the minus 2 divided by wavelengths.

02:50 And then i get 4 ,220 kelvin.

03:26 And sorry, let me pause real quick.

03:29 Okay, and then now we want to get the rate that energy is emitted from the star.

03:36 And so, i .e., we want to get the power.

03:39 And so we want to borrow some of the formulas from chapter 11.

03:43 So that tells us that power is equal to sigma e, the surface area of the star times t to the fourth.

03:53 And so this sigma, let's see, is equal to, oh, i have to pull up the chapter again.

04:02 5 .6696 times 10 of the minus 8.

04:06 Emissivity is one and a is a surface area...