One Mev positron encounters one MeV electron travelling in opposite direction. What is the wavelength of photons produced, given rest mass energy of electron or positron `=0.512MeV` ? Take `h=6.62xx10^(-34)J-s`.

Answer

`8.2xx10^(-13)m`

Explanation

The nuclear reaction is
`._(+1)e^(0)+._(-1)e^(0)to 2gamma`
Total energy before reaction =`1.512xx2MeV`
After reaction, energy of each photon,
`E=(1.512xx2)/2MeV`
`E=1.512xx1.6xx10^(-13)J=(hc)/lambda`
`lambda=(hc)/(1.512xx1.6xx10^(-13))=(6.6xx10^(-34)xx3xx10^(8))/(1.512xx1.6xx10^(-13))`
`=8.2xx10^(-13)m`