What is the speed of gold nuclei before colliding at Brookhaven collider?

[Benzoate]

Homework Statement

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?

Homework Equations

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

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) )

 

[dynamicsolo]

Homework Equations

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

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 ).

 

[Benzoate]

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 ).

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 that's part of the equation for gamma , so finding gamma would be useless.

 

[genneth]

Homework Statement

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?

Homework Equations

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

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) )

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

 

[Benzoate]

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

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

 

[robphy]

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

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" ]

 

[Benzoate]

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

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?

 

[dynamicsolo]

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.

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.

Why would I need to find gamma? I need to find the velocity, a variable that's part of the equation for gamma , so finding gamma would be useless.

?? 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.

 

[Benzoate]

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.

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

 

[robphy]

www.google.com/search?q=39400+GeV*(10%5E9+eV%2F1GeV)*(1.602e-19+J%2F1+eV)

By the way, how did you enter 10^9 in whatever you used to calculate your number?

 

[Benzoate]

www.google.com/search?q=39400+GeV*(10%5E9+eV%2F1GeV)*(1.602e-19+J%2F1+eV)

By the way, how did you enter 10^9 in whatever you used to calculate your number?

10^9 eV =1 GeV

 

[robphy]

10^9 eV =1 GeV

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

 

[Benzoate]

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

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)?

 

[dynamicsolo]

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)?

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.