To calculate the kinetic energy (KE) of an electron moving with a velocity of 0.98 times the speed of light (c), we can use the relativistic kinetic energy equation:
KE = (γ - 1) * m * c^2
Where:
- KE is the kinetic energy
- γ (gamma) is the Lorentz factor, given by γ = 1 / √(1 - v^2/c^2)
- m is the mass of the electron
- c is the speed of light
The mass of an electron (m) is approximately 9.10938356 × 10^-31 kilograms, and the speed of light (c) is approximately 299,792,458 meters per second.
Let's substitute these values into the equation and calculate the kinetic energy:
v = 0.98c γ = 1 / √(1 - (0.98)^2)
γ = 1 / √(1 - 0.9604) γ = 1 / √(0.0396) γ = 1 / 0.199
γ ≈ 5.025
KE = (5.025 - 1) * (9.10938356 × 10^-31 kg) * (299,792,458 m/s)^2
KE ≈ 4.025 * (9.10938356 × 10^-31 kg) * (299,792,458 m/s)^2
KE ≈ 4.025 * (9.10938356 × 10^-31 kg) * (8.98755179 × 10^16 m^2/s^2)
KE ≈ 3.9917 × 10^-14 Joules
Therefore, the kinetic energy of an electron moving with a velocity of 0.98 times the speed of light is approximately 3.9917 × 10^-14 Joules.