# : Perturbed Hydrogen Atom

1. Apr 30, 2004

### .cfg

Hey all. I've got a physics question due tonight (Friday) at 11pm CST and I'm stumped. Any input at all would be helpful, thanks!

A hydrogen atom, when vigorously perturbed, can emit light with a frequency of 6.16·10^14 Hz. When the same light from hydrogen atoms in a distant galaxy is observed on earth, the frequency is 5.16·10^14 Hz. Calculate the speed at which the galaxy is receding from the earth (in units of the speed of light, c).

Cheers.

2. Apr 30, 2004

### jcsd

You just need to muck around with the formula for (special, I assume with the information given) relativistic Doppler shift

$$z+1 = \sqrt{\frac{1 + \frac{v}{c}}{1- \frac{v}{c}}}$$

Where

$$z+1 = \frac{\lambda'}{\lambda}$$

3. Apr 30, 2004

### .cfg

Err.. are both velocities in the doppler shift equation equivalent or are they different? IE. would one of the velocities be zero because the receiving planet does not move?

4. Apr 30, 2004

### jcsd

There both the same.

5. Apr 30, 2004

### .cfg

I calculated lambda prime/lambda to be 1.1938. How am I to solve with two of the same variable

No matter what equation I use, it seems I keep getting an answer of 5.79*10^7.. which is pretty unrealistic seeming (and also incorrect).

Last edited: Apr 30, 2004
6. Apr 30, 2004

### jcsd

I've got the same.

Just plug z + 1 = 1.1938 into the equation (as we want v as a fraction of c anyway we can say v = v/c):

$$(z+1)^2 = \frac{1 + v}{1-v}$$

Now it's simple algerbra to re-argange it to find v

edited to add: here's a clue: a good approximation for v at non-relativistic speeds is cz, which in this case is about 0.19c, so that's the kind of figure you should be looking for (though not that exact figure as the speed is close enough to relativistic to effect the answer).

Last edited: Apr 30, 2004
7. Apr 30, 2004

### .cfg

Yea, I got that far.. it seems I'm having a brainfart with the 'simple algebra'.

EDIT: working on it with your hint
EDIT2: bleh. something's just not clicking.
5.79*10^7 is pretty close to .19c, is that telling me anything?

Last edited: Apr 30, 2004
8. Apr 30, 2004

### jcsd

Okay :

If we re-arrange the equation in my last post we get:

$$(z +1)^2(1-v) = 1 + v$$

mutiply out:

$$(z+1)^2 - v(z+1)^2 = 1 + v$$

Add v(z+1)2 - 1 to both sides:

$$(z+1)^2 -1 = v + v(z +1)^2$$

divide both sides by v:

$$\frac{(z+1)^2 - 1}{v} = (z+1)^2 + 1$$

Now just re-arrange that to get v:

$$v = \frac{(z+1)^2 -1}{(z+1)^2 + 1}$$

9. Apr 30, 2004

### .cfg

jcsd you just saved my ass. It was kind of you to help so much.

Thank you very, very much!

10. Apr 30, 2004

### jcsd

The answer you should get is about 0.17c which as you can see is pretty close to the value of 0.19c (which was obtained without the relativistic correction), though as v approaches c this approximation gets less and less accurate.