To calculate the energy of an antiparticle electron (positron) with a velocity of 0.99c, we can use the relativistic energy equation:
E = γmc^2
where E is the energy, γ (gamma) is the Lorentz factor, m is the rest mass of the particle, and c is the speed of light in a vacuum.
The Lorentz factor γ is given by:
γ = 1 / sqrt(1 - v^2/c^2)
where v is the velocity of the particle and c is the speed of light.
Given that the velocity of the antiparticle electron is 0.99c, we can substitute this value into the equation:
γ = 1 / sqrt(1 - (0.99c)^2/c^2) = 1 / sqrt(1 - 0.99^2) ≈ 7.088
The rest mass of an electron or positron is approximately 9.10938356 × 10^-31 kilograms. The speed of light is approximately 2.998 × 10^8 meters per second.
Now, we can calculate the energy:
E = γmc^2 ≈ (7.088)(9.10938356 × 10^-31 kg)(2.998 × 10^8 m/s)^2 ≈ 6.341 × 10^-13 joules (J)
So, the energy of an antiparticle electron with a velocity of 0.99c is approximately 6.341 × 10^-13 joules.