# Where do positron and electron get K.E from in Pair production

In pair production, we get a positron and electron from a photon. The photon should have 1.02Mev energy to carry out this process

When the energy of positron and electron is added, we get the energy of photon.

hf = E- + E+ = (m0c2 + K.E-) + (m0c2 + K.E+) = K.E- + K.E+ + 2m0c2

mc2 for electron or positron is = 0.51 MeV so,

hf = 2 x 0.51 + K.E- + K.E+

hf = 1.02 + K.E- + K.E+

so, where do the positron and electron get K.E from because the the sum of their energy is equal to photon.

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see answers to similar question physics.stackexchange.com/questions/56296/… . one answer has the kinematics. –  anna v Mar 9 '13 at 5:42

## 1 Answer

Consider pair production in the field of an atomic nucleus. Here, pair production leads to all the energy of the photon being converted into the rest mass of the two resulting particles and their kinetic energy. The so-called threshold energy of the photon $E_{\gamma,min}$ must therefore be at least the sum of the rest energy of the electron and positron.

Now when you write down energy and impulse conservation for this system, you have to take into account the recoil of the atomic nucleus. The threshold energy for pair production is then:

$E_{\gamma,min} = 2m_e c^2 (1 + \frac{m_e}{M_{nucleus}})$

If the photon has an energy bigger than $E_{\gamma,min}$, this excess energy is converted into the kinetic energy of the resulting positron and electron. You can often neglect the recoil from the nucleus.

The probability of pair production to take place is proportional to the atomic number of the nucleus and to the logarithm of the energy of the photon.

You're basically asking why we don't see pair production take place in vacuum.

To understand this, think about a photon that has energy $E_\gamma = 2m_e c^2$. In the CMS (center of mass system) of the created proton and electron, the sum of their impulse is equal to zero. BUT, photons have the same velocity (speed of light) in every rest frame. So the photon has the impulse $E_\gamma / c$ in the CMS of the electron and the positron. This is why the electron and the positron cannot be the only particles involved in this scattering process. You need to have the nucleus. This is why you don't see pair production happening in vacuum. THe conservation of impulse would be violated.

Griffiths particle physics includes an excellent introduction to this topic.

Btw, it's really hard to read your equations. Please use the built in Tex functionality.

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so,do you mean that if photon has energy greater than 1.02 MeV only then positron and electron will have K.E..? sorry for the equations. i don't know how to do it here... –  Rafique Mar 9 '13 at 10:40
yes. if the energy of your photon is exactly $E_{\gamma,min}$ the positron and electron will have no kinetic energy –  lomppi Mar 9 '13 at 11:00
As photon has almost negligible momentum so, if both positron and electron are at rest, momentum is conserved automatically, right? –  Rafique Mar 9 '13 at 11:08