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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2

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
 
Last edited:
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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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Likes ikihi
  • #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.
 

What is pair production?

Pair production is a phenomenon in particle physics where a high-energy photon interacts with a nucleus or an atom and produces a pair of particles, usually an electron and a positron, with opposite charges.

What is the pair production problem?

The pair production problem refers to the difficulty in understanding how a single photon can have enough energy to produce a pair of particles when the energy of the photon is less than the combined rest mass of the particles created.

What is the significance of pair production?

Pair production plays a crucial role in understanding the behavior of matter and energy at the subatomic level. It also has practical applications in fields such as medical imaging and particle accelerators.

What is the relationship between pair production and quantum mechanics?

Pair production is a result of the principles of quantum mechanics, specifically the concept of energy-mass equivalence. It also demonstrates the probabilistic nature of particle interactions at the subatomic level.

How does pair production relate to the conservation of energy and momentum?

Pair production obeys the laws of conservation of energy and momentum. Although a single photon does not have enough energy to produce a pair of particles, the energy and momentum are conserved through the interactions of the created particles with the surrounding matter.

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