Pair Production Velocity Calculation

In summary, a proton and anti-proton are created by a photon with a wavelength of 6.607×10−7 nm. The magnitude of the velocity of the newly created proton and anti-proton pair is calculated using the equations Ephoton = h ⋅ c / λ, E0 = m0 ⋅ c2, and KE = 0.5 ⋅ m ⋅ v2. After converting the energy of the photon to MeV and subtracting it from the energy of two protons (938.3 MeV), the calculated kinetic energy is 0.029 MeV, which corresponds to a velocity of 2.36 x 106 m/s. However, after using more significant
  • #1
ikihi
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Homework Statement


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A proton and anti-proton are created by a photon with wavelength λ= 6.607×10−7 nm. What is the magnitude of the velocity of the newly created proton and anti-proton pair? Note, the mass of a proton/anti-proton is mp= 1.673×10-27 kg = 938.3M MeV/c2.

Homework Equations



Ephoton = h ⋅ c / λ
E0 = m0 ⋅ c2
KE = 0.5 ⋅ m ⋅ v2

The Attempt at a Solution



Ephoton= 3.009 x 10-10 J
E0 = 1.506 x 10-10 J

KE=Ephoton - 2 ⋅ E0 / (2)
KE= 3.009 x 10-10 - 2 ⋅ 1.506 x 10-10 J / (2)
KEof either particle= -1.5 x 10-13 J
 
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  • #2
ikihi said:
E0 = 1.506 x 10-10 J
Google calculator shows 938.3 MeV=1.503E-10 Joules.
 
  • #3
if you convert the energy of the photon to Mev and subtract from it (2*938.3 Mev) it will give a positive number
i guss the problem is the mass of the proton missing a lot of digits and the speed of light as well.
like Bandersnatch said if you use the full digits of the mass and speed of light it will give 1.503*10^-10
 
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  • #4
Bandersnatch said:
Google calculator shows 938.3 MeV=1.503E-10 Joules.

I calculated E0 using E0 = m0 ⋅ c2. So is that the same as the given value?
 
  • #5
patric44 said:
if you convert the energy of the photon to Mev and subtract from it (2*938.3 Mev) it will give a positive number
i guss the problem is the mass of the proton missing a lot of digits and the speed of light as well.
like Bandersnatch said if you use the full digits of the mass and speed of light it will give 1.503*10^-10

You are right. I converted both to MeV and it came out to KE = 0.029!

0.029 MeV ---> 4.646 x10-15 J

I calculate that the speed is 2.36 x 106 m/s.

However after using more sig fig digits the answer goes to 0 m/s. Maybe this is a relativistic problem that needs a different equation?
 
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  • #6
ikihi said:
after using more sig fig digits the answer goes to 0 m/s
That seems rather unlikely. If you cannot find a mistake please post all your working.
 
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