Photon energy when its momentum equals a proton's with 10 MeV kinetic energy

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Homework Statement


What is the energy of a photon whose momentum is the same as that of a proton with a kinetic energy of 10 MeV?

Homework Equations


K = mc[tex]^{2}[/tex]([tex]\gamma[/tex]-1)
p = [tex]\gamma[/tex]mv
E = pc

The Attempt at a Solution


I figured I would go at it like this. I used the first equation listed to obtain a value for gamma and v. I put these into the second equation along with the mass of a proton. Then, I would put that value into the third equation. Is that even close?
 
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Seems ok if I understood it right. Your equations might help more.
 
As far as I can understand, the procedure you describe is correct. As long as you have done the calculations properly, you should end up with the correct answer.

Torquil
 
Ok, so I have [tex]\gamma = 1 + 1.18 \times 10^{-19}[/tex] So, um, how am I supposed to square that (so I can find beta)? My calculator won't go that far... :*)
 
use an online calculator
 
or any computer program will do it
 
or you can just square the 1.18 and then double 19
 
dacruick said:
use an online calculator
They all give me 1
 
the answer is 1.3924e-38
 
so for next time this is what you will do. you should square 1.18. then you should double the exponent. and voila
 
dacruick said:
so for next time this is what you will do. you should square 1.18. then you should double the exponent. and voila
You're misreading the problem. If it was just 1.18e-19, I wouldn't have a problem. I need to square 1+1.18e-19. BTW, I found a calculator that will do it. Google can be tricksy sometimes...
 
if [itex]\gamma=1+1.18\times10^{-19}[/itex] then you should just approximate [itex]\gamma\simeq1[/itex].

Also, use www.wolframalpha.com It's like having Mathematica available to you with 0 cost
 
Your value for [itex]\gamma[/itex] is way too small. The proton's kinetic energy is about 1% of the rest energy, so [itex]\gamma[/itex] should be about 1.01.
 
lockedup said:

Homework Statement


What is the energy of a photon whose momentum is the same as that of a proton with a kinetic energy of 10 MeV?



Homework Equations


K = mc[tex]^{2}[/tex]([tex]\gamma[/tex]-1)
p = [tex]\gamma[/tex]mv
E = pc



The Attempt at a Solution


I figured I would go at it like this. I used the first equation listed to obtain a value for gamma and v. I put these into the second equation along with the mass of a proton. Then, I would put that value into the third equation. Is that even close?

The speed of light = 299,792,458 m/s;
Proton mass: [tex]1.67262158*10^{-27} kg[/tex];
[tex]K= 1.602176487*10^{−15} J[/tex].

So with a simple mind-based calculation you at least can get that the Lorentz factor is around 1. The following shows its exact value up to 9 decimal digits:

[tex]\gamma=1.000010658[/tex].

AB
 
This is such a complicated way to do this problem. The energy of a proton is its kinetic energy plus its mass, and that squared is p^2 + m^2 (with c=1). Find p. E(photon) = p. Done.