Calculating the Momentum of a Proton with a Kinetic Energy of 750MeV

[UAPhys03]

Homework Statement

What is the momentum of a proton with a kinetic energy of 750MeV?

Homework Equations

E = m₀c² + K
E² = p²c² + m₀c²
Mp = 1.67 x 10^-21

The Attempt at a Solution

I think what's confusing me is K = 750MeV, but I tried to solve for p

First, I converted 750MeV to 750x10⁶ eV

Then solved for E in the first equation
E = 750x10⁶ eV

Then used that to solve for p in the second equation
But i get a wrong answer.

The correct answer is 1404MeV/c

 

[hage567]

In your second equation, m_oc^2 needs to be squared. So it should be E^2 = p^2c^2 + (m_oc^2)^2

You also need to express the proton mass in electronvolts. Are you doing that?

 

[Redbelly98]

Homework Statement

What is the momentum of a proton with a kinetic energy of 750MeV?

Homework Equations

E = m₀c² + K
E² = p²c² + m₀c²
Mp = 1.67 x 10^-21

The Attempt at a Solution

I think what's confusing me is K = 750MeV, but I tried to solve for p

First, I converted 750MeV to 750x10⁶ eV

Then solved for E in the first equation
E = 750x10⁶ eV

In other words, you get that E = K. But that's only true when m₀=0, which is not the case for a proton.

You did mean to say proton, and not photon?

 

[UAPhys03]

Hey,

So I fixed the equation and figured it out

E = m₀c² + KE

m₀c² = 1.5x10^-10j = 939.4MeV

E = 939 + 750 = 1689MeV

E² = p²c² + m₀²c²

1689² = p²c² + 939²

p²c² = 191000

pc = 1404MeV

p = 1404MeV/c

Thanks!

 

[apunisheriii]

is it okay to solve like this??

1/2(Mp)v²=K
1/2(p²)/Mp=K
p=square root of 2K(Mp)
p=6.34x10^-19 kg m/s

 

[Andrew Mason]

Hey,

So I fixed the equation and figured it out

E = m₀c² + KE

m₀c² = 1.5x10^-10j = 939.4MeV

E = 939 + 750 = 1689MeV

E² = p²c² + m₀²c²

1689² = p²c² + 939²

p²c² = 191000

pc = 1404MeV

p = 1404MeV/c

Thanks!

Your arithmetic is not correct. It should be: p²c² = 1,970,000

AM

 

[Andrew Mason]

is it okay to solve like this??

1/2(Mp)v²=K
1/2(p²)/Mp=K
p=square root of 2K(Mp)
p=6.34x10^-19 kg m/s

Not for relativistic speeds. A proton with 750 Mev of kinetic energy is moving at relativistic speed. At non-relativistic speed you can use:

\text{KE} = \frac{1}{2}mv^2 = \frac{p^2}{2m}

p = \sqrt{2m\text{KE}}

You had it right if you were using Mp for proton mass and p for momentum, but it is confusing.

AM

 

[apunisheriii]

Not for relativistic speeds. A proton with 750 Mev of kinetic energy is moving at relativistic speed. At non-relativistic speed you can use:

\text{KE} = \frac{1}{2}mv^2 = \frac{p^2}{2m}

p = \sqrt{2m\text{KE}}

You had it right if you were using Mp for proton mass and p for momentum, but it is confusing.

AM

How can i know whether it is in non-relativistic or relativistic speed?

 

[Redbelly98]

For non-relativistic speeds, KE has to be a lot less than mc². Equivalently, this means the speed is a lot less than c.

In this problem, KE is 750 MeV. That is comparable to mc² = 940 MeV, so it is relativistic in this case.

 

[UAPhys03]

So for relativistic speed wouldn't you just substitute that equation,

K = 1/2((lorenz factor) - 1)mv²

substitute for p

K/((lorenz factor) - 1) = p²/2m

and you could figure out how fast the proton is moving given the kinetic energy?

 

[Redbelly98]

No, the relation is

K = ( (Lorenz factor) - 1 ) m c²

Given K, solve for the Lorenz factor, from which you can get v.

 

[apunisheriii]

So can i use E=hf to calculate the momentum as it's relativistic??

 

[Redbelly98]

Why would you think that? Momentum does not appear anywhere in your equation ?

 

[apunisheriii]

i thought the wavelength in the equation E=hc/lambda is the same as the de Broglie's wavelength...
is the de Broglie's wavelength of a electron with non-relativistic speed couldn't be sub into the E=hf equation??

 

[Redbelly98]

Sometimes, not always. There actually 3 "speed regimes", not 2, to think about.

1. Non-relativistic: K is much less than mc²
2. Moderately relativistic: K is comparable to mc²
3. Extremely relativistic: K is much larger than mc²

That equation, E = hf = hc/lambda = pc , is valid in the extremely relativistic case. Another way of seeing this is from two equations that are true in all 3 cases:

E² = (pc)² + (mc²)²
and
E = K + mc²

In case 3 (and only then) we can neglect the mc² terms because it is much smaller than the other quantities, and we get

E = K = pc (and also E = hc/lambda = hf)

Hope that helps.

 

[apunisheriii]

o,
finally i understood!
thnx ya!