# Homework Help: Relavistic energy problem.

1. Sep 26, 2007

### Benzoate

1. The problem statement, all variables and given/known data
The relavtivistic Heavy Ion collider at brookhaven is colliding fully ionized gold(Au) nuclei accelerated to an energy of 200 GeV per nucleon. each Au nucleuss contains 197 nucleons . a)what is the speed of each Au just before the collision?

2. Relevant equations

E=2mc^2/sqrt(1-u^2/c^2)

3. The attempt at a solution

Energy= (200 GeV)(10^9 eV/GeV)(1.609e-19 Joules/1 eV)= 6.34e-5 joules
m(proton)= 1.676362e-27 kg =1 u => 197 u = 3.302e-25 kg

I got my mass and my Energy values, with these values, I should now be able to find u.

E=2mc^2/sqrt(1-u^2/c^2) => sqrt(1-u^2/c^2) =2mc^2/E => 1-u^2/c^2 = 4m^2*c^4/(E^2) => 1-4m^2*c^4/(E^2) = u^2/c^2 => c^2* (1-4m^2*c^4/(E^2) )=u^2 => u=sqrt(c^2* (1-4m^2*c^4/(E^2) )

2. Sep 26, 2007

### dynamicsolo

Where does this come from? I believe this problem is not looking at the imminent collision, but simply the velocity of one of the gold nuclei.

The energy of each nucleus, and thus each nucleon in the nucleus, is given as 200 GeV. The energy is E = (gamma)(mc^2). It will be helpful here to use the rest mass-energy of a nucleon in electron-volts (0.9383 GeV for a proton; 0.9396 GeV for a neutron); it will be good enough to take the rest mass-energy per nucleon as 0.939 GeV.

So the ratio of the given energy to the rest energy will give you gamma. (I'm not clear from the problem statement whether the "energy given" is the kinetic energy or the total energy, but since K = (gamma - 1)(mc^2), with gamma as high as it is, there won't be much difference in your answer. You can then use the equation for gamma to tell you beta = ( u/c ).

3. Sep 27, 2007

### Benzoate

You can find the energy of the gold nucleu by simply multiplying the number of nucleons , in this case, 197 * the accelerated energy per nucleon, in my case, the E being equal to 200 GeV/nucleon. In order t o find mass you just multiply the mass of a proton * the mass numbeer of the Au atom, in my case its mass being 197 u. Why would I need to find gamma? I need to find the velocity, a variable thats part of the equation for gamma , so finding gamma would be useless.

4. Sep 27, 2007

### genneth

So what's the issue? Does the answer come out wrong?

5. Sep 27, 2007

### Benzoate

I'm not sure how to convert my momentum unit kg *m/s to GeV/c

6. Sep 27, 2007

### robphy

What is a "GeV"? Can you express that in some more familiar unit? If not, you can google it.
[Edit: you can google "1 GeV" ]

Last edited: Sep 27, 2007
7. Sep 27, 2007

### Benzoate

Thanks. I don't have the correct momenta. relavistic momentum= m*u*gamma. m =197 u*(1.673e-27 kg/u)=3.296e-25 kg
E=197 nucleons*(200 GeV/nucleon) 39400 GeV*(10^9 eV/1GeV)*(1.602e-19 J/1 eV)= 6.312e-5 joules. E=mc^2*gamma => E=mc^2*1/sqrt(1-u^2/c^2)=> sqrt(1-u^2/c^2)=mc^2/E=> 1-u^2/c^2=m^2*c^4/E^2 => u^2/c^2 =1-(2.21e-7), m^2*c^4/E^2=2.21e-7 a u^2/c^2 = .999999779 => u^2 = 8.999998012m^2/s^2 => u =299999966.9m/s . Now that I have my mass and velocity , I can now find my momentum. p=mu*gamma= (3.296e-25 kg)(299999966.9m/s)(1/sqrt(1-(299999966.9m/s)^2/(9e16 m^2/s^2))=2.1049e-13 kg*m/s

(1 gigaelectron volt) / the speed of light = 5.34428542 × 10^-19 m kg / s therefore , 2.1049e-13 kg *m/s *(1 GeV/c/(5.34428*10e-19 kg*m/s)= 393860.352 GeV

According to my textbook , my momentum is wrong. . Perhaps there is something wrong with my math?

Last edited: Sep 27, 2007
8. Sep 27, 2007

### dynamicsolo

True, but since every nucleon is traveling together in the nucleus, and you just want to know the velocity in part (a), you can just use the 200 GeV/nucleon.

?? You find gamma because the problem is pretty much telling you what it is. It is easier to then find u from (gamma) than to try to extract it from the mechanics first. Besides, for the part of the problem where you calculate the momentum, don't you need u and gamma?

I am finding a velocity more like u = 0.9999875 c. I believe this

E=197 nucleons*(200 GeV/nucleon) = 39400 GeV*(10^9 eV/1GeV)*(1.602e-19 J/1 eV)= 6.312e-5 joules

should give 6.312e-6 joules.

Last edited: Sep 27, 2007
9. Sep 27, 2007

### Benzoate

no , I'm getting 6.312e-5 Joules, not 6.312e-6 joules

10. Sep 27, 2007

### robphy

11. Sep 27, 2007

### Benzoate

12. Sep 27, 2007

### robphy

What I mean is....
did you enter "10^9" or "1e9" or (the incorrect) "10e9"?

13. Sep 27, 2007

### Benzoate

I have another question about finding the mass of Au particle.to calculate the mass I just multiply atom mass o Au atom * total mass of proton. m=(197 u)*(938.57 MeV/c^2/(u)) and E= 197 nucleons*(200 GeV/nucleon)?

14. Sep 27, 2007

### dynamicsolo

That would be close enough for the purpose of problems such as this one. A more precise value for the nuclear mass of Au-197 would be (79 x Mp) + (118 x Mn) - (nuclear binding energy), but that's needlessly fussy for this.

BTW, u is not the proton's mass, but rather the atomic mass unit. It is currently based on the nuclear mass of C-12, so u = [ (6 x Mp) + (6 x Mn) - (binding energy)]/12. If you look up values for the physical constants at the NIST site, for example, you'll see that Mp, Mn, and u are all pretty similar, but not exactly the same.