00:01
There, so for this problem, we are told that doppler shift of a hydrogen spectrum lined.
00:07
One of the most prominent spectral lines of hydrogen is the alpha hydrogen lined, a bright red line with a wavelength that is given, and that wavelength is equal to 6506 .1 times 10 to minus 9 meters.
00:30
So for part a of this problem, we are asked about what is the expected wavelength of the hydrogen alpha line for a start that is receding with speed.
00:44
And the speed is given and that speed is equal to 3 ,000 kilometers per second.
00:59
That we can also write as 3 times 10 to 6 meters per second.
01:11
So with this, we are going to first, we note that light from a resident source is shifted toward the red, longer wavelength, and the light from an approaching source is shifted toward the blue.
01:29
So in this case, we're going to have that.
01:33
The velocity prime of the, this moving object is the value given for the velocity times one minus the velocity divided by the speed of light, divided by one plus the velocity divided by the speed of light.
01:50
And then we know from this that from the information given that the ratio between the velocity and the speed of light that is equal to three times 10 to the six meters per second divided by three times 10 to the eight, meters per second for the speed of light, we obtain 0 .01.
02:16
So using this equation, we know that the wavelength prime is equal to the speed of light divided by the speed prime.
02:29
So we will have that this is the wavelength times one plus this ratio, one minus this ratio.
02:39
In here we have tate inverse.
02:42
Of the previous expression.
02:45
We can approximate this to the wavelength times 1 plus 2 times the speed divided by the speed of light.
02:59
That we can also approximate and all of these approximations are valid because we know that the speed of this is much less than the speed of light.
03:11
And then we can approximate this to the wavelength times 1 plus this.
03:18
So from this we obtain that the wavelength should be 1 .01 times the initial wavelength given.
03:26
So we know now that the change in wavelength which is equal to this which is equal to 0 .01 times the initial wavelength is equal to 6 .6 times 10 to minus 9 meters.
03:45
So since we want the wave.
03:47
Prime, we just solve for this and we will obtain that this is 6506 .1, the value that is given, plus the value that we obtained from this, which is 6 .6 times 10 to the minus 9 meters.
04:05
And this times 10 to the minus 9.
04:10
So with this, we obtain that the wavelength in this case should be equal.
04:16
This is the aspect of wavelength.
04:21
From this rescind star should be 662 .7 times 10 to the minus 9 meters.
04:29
So that's a solution for part a of this problem.
04:35
Now for part b, we are told that the lines of hydrogen measure on air from opposite ends of the sun's equator deferred in wavelength by 9 times 10 to the minus 12.
04:55
Meters.
04:57
So the question is assuming that the effort is caused by rotation on the sun, we need to find the period of rotation.
05:06
And given that the diameter of the sun is given and that is 1 .4 times 10 to the 6 kilometers...