0:00
Hi there.
00:01
So for this problem, we are told that sirius b, the brightest star in the sky, is serious.
00:05
The dark star.
00:06
It is actually a binary system of two stars, the smaller one, series b, being a white dwarf.
00:12
And spectral analysis of series b indicates that its surface temperature is 24 ,000 kelvin.
00:19
So we're given that temperature, 24 ,000 kelvin, and that it radiates and an energy.
00:31
At a total rate that is also given, and that is equal to a value, let's call it that p, and that is equal to 1 times 10 to the 25 watts.
00:50
So assume that it behaves like an ideal black body.
00:53
So for part a of this problem, we are asked about what is the total radiated intensity of serious b? so to obtain intensity, we use the usual equation that obeyed the stefan boltzmann law that states that this is equal to the cittma constant, the stephen bolsman constant times the temperature to the four.
01:15
So we substitute those values.
01:16
This constant is equal to 5 .67 times 10 to the minus 8 in units of watts per meter square per kelvin to the 4.
01:26
And these times the temperature that we are given, 24 ,000 kelvin and that to the 4.
01:32
So from this, we obtain an intensity of 1 .9 times 10 to the 10 watts per meter square.
01:41
So that's a solution for part a of this problem.
01:46
Now, for part b, we are asked about what is the peak intensity wavelength? is this wavelength visible to humans? now, to obtain this, we use the equation that states that the viewing's law, the law states that the maximum wavelength peak wavelength is equal to a value, a constant value that is 0 .00 to 190 meters times kelvin.
02:16
And this divided by the temperature that we are given, which is the 24 ,000 kelvin.
02:22
So from this, we obtain a value of 1 .2 times 10 to the minus 7 meters.
02:28
So that will give us a value of 20 nanometers.
02:35
Oh, sorry, just simply 120 nanometers.
02:45
120 nanometers.
02:46
So we know that the visible light spectrum is between around 400 nanometers and 700 nanometers.
02:56
So this is less than that value.
02:58
So we can say that this is not visible, is not visible, since this wavelength is less than 400 nanometers.
03:11
So we now pass to the last question in here, which is question c.
03:18
Oh, sorry, no, is the next question, question c, which is, what is the radius of series b? now, to obtain the radius of series b, we use the following equation that states that, power is equal to the surface area times intensity.
03:36
Now, the surface area is defined as four times pi times the radius to the square, the radius of the star to the square.
03:43
So what we need to do is to simply solve for the radius.
03:46
So we will have that the radius is equal to the power that we are given divided by four times pi times the intensity, and we need to take the square root of all.
04:03
Of this.
04:04
So using the values that we are given, remember that the power is given, that is 1 times 10 to the 25 watts.
04:11
So let me substitute that in here...