Question

The minimum amount of energy released in annihilation of electron-positron is.

• A. $1.02MeV$
• B. $0.58MeV$
• C. $185MeV$
• D. $200MeV$

Answer 1

The correct option is A $1.02MeV$

Energy released during the annihilation of $e^{-},e^{+}$ pair is,

$E=2m_c{C}^2$

$E=2\times 9.1\times {10}^{-31}\times {(3\times {10}^8)}^{2}J$

$E=\frac{2\times 9.1\times 9\times {10}^{-15}}{1.6\times {10}^{-13}}MeV$

$E=1.02MeV$