Question

what is the minimum amount of energy released in the annihilation of an electron position pair.

No spam

Answer 1

What is the minimum energy released in annihilation of electron positron pair?

Both positron and electron are of equal mass i.e about 9.1*10^-31 kg .

The speed of light is about 3*10^8 m/s, and as per Einstein's theory E=m*c*c .

Total mass = 2*9.1*10^-31 and c*c = (3*10^8)^2 = 9*10^16.

So, energy released = 2*9.1*10^-31*9*10^16 = about 1.637*10^-13 joules, or about 1.02 MeV.

Answer 2

The amount of energy is 1.02mev

Given,

An electron-positron pair.

To Find,

The minimum amount of energy released in the annihilation.

Solution,

An electron-positron pair is there.

Need to find the minimum amount of energy released in the annihilation.

Electron and positron have a rest mass of the same m₀ = 9.1 × $10^{-31}$kg.

The mass is converted into the energy form during annihilation,

By using the Einstein equation,

E = Δ m$c^{2}$

Here, one pair is present so, Δm will be 2m₀.

E = 2m₀$c^{2}$

E= 2 × 9.1 ×$10^{-31}$×$c^{2}$

C is the light's speed.

E = 2 × 9.1 × $10^{-31}$ × $(3*10^{8)} ^{2}$

E =  2 × $9.1 *10^{-31} * 9* 10^{16}$

1mev = 1.602 * $10^{-13}$J

E = $\frac{2*9.1 * 10^{-31} * 9 *10^{16} }{1.602*10^{-13} }$

E = 102mev.

Hence, 1.02mev is the amount of energy released by the electron-positron pair in the annihilation.

#SPJ3