1.

To produce an electron-position pair, the minimum energy of `gamma`-ray photon must beA. `1.02keV`B. `1.02MeV`C. `1.02BeV`D. `1.02eV`

Answer» Correct Answer - B
Rest energy of electron `= (9.1xx10^(-31)xx(3xx10^(8))^(2))/(1.6xx10^(-13))MeV`
`= 0.510 MeV`
`gamma rarr e^(+) +e^(-)`
Energy of photon `= 2xx0.51 = 1.02MeV`


Discussion

No Comment Found

Related InterviewSolutions