Calculating Speed of Protons in a Linear Accelerator: 530 MeV Kinetic Energy

In summary, using the equation E = KE + E0 and recognizing that 1MeV = 10^6 eV, the speed of protons in a proton linear accelerator with a kinetic energy of 530 MeV is found to be approximately 2.307E8 C. To find the mass of the proton (not rest mass), one could use the equation E = mγc^2 and solve for m.
  • #1
ilovejava
21
0

Homework Statement


In a proton linear accelerator, protons are accelerated to have a kinetic energy of 530 MeV. What is the speed of these protons? (The rest mass of a proton is 1.67 × 10-27 kg.)

Homework Equations


E0=m0c^2
E=E0/√1-(v^2)/(c^2)
E = KE + E0

The Attempt at a Solution


Recognize that 1eV = 1.602E-19 Joules and 1MeV = 10E6 eV. I converted 530MeV to joules and it turned out to be 8.49E-11. I found the total energy using E = KE + E0 to be 2.35E-10. I then solved for velocity using E=E0/√1-(v^2)/(c^2) and it turned out to be 2.306E8 C
 
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  • #2
You did something wrong. What was your value for E0 and E/E0 ?
 
  • #3
ilovejava said:
I converted 530MeV to joules and it turned out to be 8.49E-11.

Surely you ought to do this problem in MeV? That's why these units were invented!
 
  • #4
phyzguy said:
You did something wrong. What was your value for E0 and E/E0 ?
I found E0 = 1.503E-10 and E/E0 = 0.63957
 
  • #5
ilovejava said:
I found E0 = 1.503E-10 and E/E0 = 0.63957

The proton can't have less energy than its rest energy.

I think you also need to try to move away from the plug-and-chug mentality. You should be looking to express speed in terms of energy and do one calculation at the end.
 
  • #6
PeroK said:
The proton can't have less energy than its rest energy.

I think you also need to try to move away from the plug-and-chug mentality. You should be looking to express speed in terms of energy and do one calculation at the end.

The rest energy of the proton (E0) is 1.503E-10 and the total energy of the proton (E = KE + E0) is 2.35E-10, so I don't know what you mean when the proton can't have less energy than its rest energy.
 
  • #7
ilovejava said:
The rest energy of the proton (E0) is 1.503E-10 and the total energy of the proton (E = KE + E0) is 2.35E-10, so I don't know what you mean when the proton can't have less energy than its rest energy.

You had:

ilovejava said:
I found E0 = 1.503E-10 and E/E0 = 0.63957

So, ##E = 0.6 E_0## is less than the rest energy.

Don't you see how you are not being defeated by the physics, but the mass of complicated numbers. You must do this problem in MeV. It will be so much simpler.
 
  • #8
ilovejava said:
The rest energy of the proton (E0) is 1.503E-10 and the total energy of the proton (E = KE + E0) is 2.35E-10, so I don't know what you mean when the proton can't have less energy than its rest energy.

If E > E0, then how can E/E0 be less than 1?
 
  • #9
PeroK said:
You had:
So, ##E = 0.6 E_0## is less than the rest energy.

Don't you see how you are not being defeated by the physics, but the mass of complicated numbers. You must do this problem in MeV. It will be so much simpler.
I apologize E0/E = 0.63957. It's funny because our professor told us to convert to joules first, so it makes it easier working with the units. If I did do the problem in MeV the approach would be the same right?
 
  • #10
ilovejava said:
I apologize E0/E = 0.63957. It's funny because our professor told us to convert to joules first, so it makes it easier working with the units. If I did do the problem in MeV the approach would be the same right?

Let me explain the advantage of MeV. The mass of a proton is ##938 MeV/c^2##. That means that the rest energy of the proton is:

##E_0 = mc^2 = 938MeV##

That's part of the advantage: the ##c^2## cancels out.

The other suggestion is that if you start from:

##E = \gamma mc^2 = \gamma E_0##, then (as a simple exercise you can show):

##\frac{v}{c} = \sqrt{1 - (\frac{E_0}{E})^2}##

Now, you just need one calculation and you don't have to struggle with numbers to 6 decimal places everywhere.
 
  • #11
phyzguy said:
If E > E0, then how can E/E0 be less than 1?
I apologize, E/E0 yields 1.56 and it was E0/E that gives 0.639
 
  • #12
PeroK said:
vc=√1−(E0E)2vc=1−(E0E)2\frac{v}{c} = \sqrt{1 - (\frac{E_0}{E})^2}

This is what I have been using and and I get 2.306E8 C as my final answer if I used joules. Next, keeping MeV is neat because it you get to work with smaller numbers, and using the numbers provided in the question I would get 2.307E8 C as my final answer. As a follow up if I wanted to find the mass of the proton (not rest mass), how would I go about doing that?

Thanks once again.
 
  • #13
ilovejava said:
This is what I have been using and and I get 2.306E8 C as my final answer if I used joules. Next, keeping MeV is neat because it you get to work with smaller numbers, and using the numbers provided in the question I would get 2.307E8 C as my final answer. As a follow up if I wanted to find the mass of the proton (not rest mass), how would I go about doing that?

Thanks once again.

I don't see the problem. ##E_0 = 938 MeV## and ##E = 530 + 938 MeV## and it's the same ratio as you eventually got using Joules. Joules are definitely inappropriate for atomic energies.

The mass of the proton doesn't change. It's rest mass is its mass. Don't tell me your professor wants you to use "relativistic mass"?
 
  • #14
PeroK said:
I don't see the problem. ##E_0 = 938 MeV## and ##E = 530 + 938 MeV## and it's the same ratio as you eventually got using Joules. Joules are definitely inappropriate for atomic energies.

The mass of the proton doesn't change. It's rest mass is its mass. Don't tell me your professor wants you to use "relativistic mass"?
Haha. You guessed correctly, he wants us to use relativistic mass.
 
  • #16
PeroK said:
That's just ##m_R = \gamma m##. But, as you may know, few people use it for reasons explained on our FAQ:

https://www.physicsforums.com/insights/what-is-relativistic-mass-and-why-it-is-not-used-much/

As a follow up question. If I want to find the mass of this proton how would I go about doing that. There is m = m0 / √1-(v2/c2). The speed of the proton turned out to be 2.306E8 C, but if I plugged that into the formula I would get a negative answer in the denominator. So, how can I find the mass of proton in order to find its momentum?
 
  • #17
You're still doing something waaaay wrong. If [itex] \frac{E_0}{E} = \sqrt{1-\frac{v^2}{c^2}} = 0.64 [/itex], then [itex] 1-\frac{v^2}{c^2} = 0.8[/itex]. This will give v/c between 0 and 1, not v = 2.3E8 c.
 
  • #18
phyzguy said:
You're still doing something waaaay wrong. If [itex] \frac{E_0}{E} = \sqrt{1-\frac{v^2}{c^2}} = 0.64 [/itex], then [itex] 1-\frac{v^2}{c^2} = 0.8[/itex]. This will give v/c between 0 and 1, not v = 2.3E8 c.

E = E0 + KE

PeroK said:
The mass of a proton is 938MeV/c2938MeV/c2938 MeV/c^2. That means that the rest energy of the proton is:

E0=mc2=938MeVE0=mc2=938MeVE_0 = mc^2 = 938MeV

Therefore,

E = E0 + KE becomes
E = 938MeV + 530MeV
E = 1468MeV

Now, let's use the equation E = E0 / √1-(v2/c2)

E√1-(v2/c2) = E0

√1-(v2/c2) = E0 / E

1 - (v2/c2) = (E0/E)2

v2/c2 = 1 - (E0/E)2

PeroK said:
vc=√1−(E0E)2vc=1−(E0E)2

Solve for V and you will get 2.3E8
 
  • #19
Sorry, I mis-typed it. So you mean 2.3E8 m/sec = 0.77 c? This looks correct. You kept typing 2.3E8 c, which is obviously impossible.
 
  • #20
ilovejava said:
E = E0 + KE
Therefore,

E = E0 + KE becomes
E = 938MeV + 530MeV
E = 1468MeV

Now, let's use the equation E = E0 / √1-(v2/c2)

E√1-(v2/c2) = E0

√1-(v2/c2) = E0 / E

1 - (v2/c2) = (E0/E)2

v2/c2 = 1 - (E0/E)2
Solve for V and you will get 2.3E8

Yes, although you need the units of ##m/s##. Often you can leave the speed as a fraction of ##c##. In this case: ##v = 0.77c##.
 
  • #21
phyzguy said:
Sorry, I mis-typed it. So you mean 2.3E8 m/sec = 0.77 c? This looks correct. You kept typing 2.3E8 c, which is obviously impossible.

Yes, my fault should've included units. Sorry for the confusion.
 

Related to Calculating Speed of Protons in a Linear Accelerator: 530 MeV Kinetic Energy

1. How is the speed of protons in a linear accelerator calculated?

The speed of protons in a linear accelerator is calculated using the equation v = √(2KE/m), where v is the speed, KE is the kinetic energy, and m is the mass of the proton.

2. What is the significance of 530 MeV kinetic energy in the calculation?

530 MeV (mega electron volts) is the amount of energy that the protons have when they enter the linear accelerator. This energy is converted into kinetic energy as the protons are accelerated through the linear accelerator.

3. How does a linear accelerator accelerate protons?

A linear accelerator uses electromagnetic fields to accelerate protons. As the protons pass through the accelerator, they are pushed by alternating electric fields and pulled by constant magnetic fields, resulting in an increase in speed.

4. How accurate is the calculation of the speed of protons in a linear accelerator?

The calculation of the speed of protons in a linear accelerator is highly accurate. However, it may vary slightly due to factors such as the precision of the equipment and external influences like air resistance.

5. Can the speed of protons in a linear accelerator be controlled?

Yes, the speed of protons in a linear accelerator can be controlled by adjusting the strength and timing of the electric and magnetic fields. This allows for precise acceleration and control of the protons' speed.

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