Calculating v/c of a proton and electron

[chenying]

Homework Statement

Calculate the speed parameter v/c of a particle with a momentum of 12.5 MeV/c if the paricle is

proton

Homework Equations

E_{}2 = m_{}2c_{}4 + p_{}2c_{}2


3. The Attempt at a Solution

Btw, the numbers above are powers.

Ok, so i followed what my teacher did:


E_{}2 = m_{}2c_{}4 + p_{}2c_{}2

m is so small for a proton that it is 0, so that is neglected


E_{}2 = p_{}2c_{}2

E = p*c = mv\gamma * c

Multiply the right side by c/c to get:

E = p*c = mv/c\gamma * c^2

then my teacher moved mc^2 to the other side to get

E/mc^2 = v/c*\gamma

Finding the value of E/mc^2, and solving for v/c, I got the value of 1, which makes sense.

That is, however, not the right answer.

Can anyone tell me what I did wrong?

 

[turin]

... a particle with a momentum of 12.5 MeV/c ...

m is so small for a proton that it is 0, so that is neglected

No. Did you mean momentum = 12.5 GeV/c? What is the mass of the proton (in MeV/c²)?

 

[chenying]

No, I meant 12. MeV.

and I do not know what you mean by the mass of a proton in MeV/c^2

 

[turin]

In this case, the approximation that m_p=0 is quite bad. Look up "proton rest energy". The first website that google found for me gave the the mass of the proton in MeV/c². Also, the Particle Data Group is an authoritative reference for this kind of thing, and they also have a website.

 

[chenying]

In this case, the approximation that m_p=0 is quite bad. Look up "proton rest energy". The first website that google found for me gave the the mass of the proton in MeV/c². Also, the Particle Data Group is an authoritative reference for this kind of thing, and they also have a website.

Well, the value I found for MeV/c² is 938.27.

So I multiplied by c to get MeV/c, which is 2.814811e11

Why is the assumption that mass of proton is 0 bad? I did the same process for the electron and I got 1, which was correct.

 

[w3390]

First of all, it should be a glaring red flag if you get the ratio of the speed of a proton to the speed of light to be 1. This is saying that the proton is traveling at the speed of light which is impossible. When I went through and solved for the velocity of the proton, I found it to be much smaller than the speed of light.

On to the problem: you have a problem that is asking for the ratio of the speed of the proton to the speed of light. There are two main unknowns, the relativistic energy of the proton and the speed of the proton. Therefore, I would suggest using two equations. I personally would use these two equations:

1. (u/c)=(pc)/E, where u is the velocity of the proton
2. E=(mc^2)/(sqrt(1-(u^2/c^2)))

Plugging equation 2 into equation 1 eliminates the energy and you can do some algebra to solve for u. By the way, when doing this algebra, do not plug any numbers in until you have simplified completely. If you do things correctly, you will never have to plug in the value of c.

Responding to your most recent post, do not multiply the 938.27 by c. You will see that the c^2 will cancel and you won't have to worry about. Try out what I have just said and see if it makes sense.

 

[chenying]

Thanks, that helped a lot.

But where did you get equation 1? Is it a ratio or was it derived?

 

[chenying]

does the p in equation 1 represent?

 

[w3390]

Equation 1 can be found by multiplying the equation for relativistic momentum by c^2 and comparing it to the equation for relativistic energy. The p in the equation represents momentum.

 

[chenying]

So then what is the p in the equation?

And I did the algebra, and tried to solve for v, but i keep running into trouble.

I keep getting a value of 0 whenever I use the algebra...

 

[w3390]

Once again, the p in the equation is momentum. I cannot see what you have done to get zero. Maybe post the steps leading up to the problem and I may be able to tell you where you went wrong.

 

[chenying]

\frac{v}{c} = \frac{pc}{E}

\frac{v}{c} = \frac{mv\gamma}{mc^2\gamma}

in which case becomes

\frac{1}{} = \frac{1}{c}

i realized originally i made a mistake but now i got something different...

 

[w3390]

Okay, first of all just leave the p as it is throughout all of the algebra. Since you know what p is you can just leave it as p in your equations. So instead of mv(gamma), it should be pc. Then you can solve for v by multiplying by c. This gives you v=(pc^2)/E. Take it from here without substituting anything in for p. Also, write gamma as
1/(sqrt(1-(v^2/c^2))). You should then manipulate the equation until you are able to easily square out the square root. Then solve for v.

 

[turin]

Well, the value I found for MeV/c² is 938.27.

So I multiplied by c to get MeV/c, which is 2.814811e11

Why is the assumption that mass of proton is 0 bad? I did the same process for the electron and I got 1, which was correct.

The MeV and c are unit factors, not physical quantities. (Well, they are also physical quantities, but we are using them as units in this problem). You cannot compare MeV/c² to MeV/c any more than you can compare, say, velocity and acceleration. The numerical value of 938.27 does not change, regardless of multiplication by c, because c is already included in the unit. So, the thing to do is to compare the momentum to mc.

mc = (938.27 MeV/c²)c = 938.27 MeV/c

p = 12.5 MeV/c

So, p << mc, which means that the mass is nonnegligible. You cannot assume m=0. In fact, doing so is tantamount to assuming v=c, period. Whoever told you to use this approximation in order to solve the problem was just plain wrong. Use:

pc = γβmc²

where γ=1/sqrt(1-β²), and β=v/c.

 

[chenying]

um... I solved for v and got 3e8..... I have no idea what I'm doing wrong...

sorry if I am frustrating you btw.

 

[chenying]

Turin:

I did it your way and I also got that v = 3e8...

 

[turin]

We cannot help you effectively if you don't show us your work. Let's try an intermediate calculation: determine β². Also, to give you a clue about what's going on, you should realize that the numerical value of β is actually the numerical value of speed in units of c.

BTW, don't worry about frustrating me. If frustration were a problem, I would have stopped coming here years ago. :)

edit:
Oh, I just realized that β is precisely what the problem asks you to calculate. Anyway, though, I hope that you will begin to realize that it's all just a matter of your unit system, and using c as a unit is particularly convenient in relativity problems.