Question

What is the energy of each of the two photons produced in an electron-positron annihilation?

Answer 1

Answer:

511797.76 eV

Explanation:

Mass of electron = Mass of positron = 9.11×10⁻³¹ kg

Mass of annihilation

Δm = Mass of electron + Mass of positron

⇒Δm = 9.11×10⁻³¹+9.11×10⁻³¹

From Einstein's Equation

$E=\Delta c^2\\\Rightarrow E=18.22\times 10^{-31}\times (3\times 10^8)^2\ J$

1 eV = 1.602 × 10⁻¹⁹ J

$\frac{18.22\times 10^{-31}\times (3\times 10^8)^2}{1.602\times 10^{-19}}=1023595.51\ eV$

Energy of each photon is

$\frac{1023595.51}{2}=511797.76\ eV$

∴ Energy of each of the two photons produced in an electron-positron annihilation is 511797.76 eV