A relativistic proton has a momentum of 1.0×10^-17 kg ∙ m/s and a rest energy of 0.15 nJ. What is the kinetic energy of
this proton? (c = 3.00 × 108 m/s, mass of proton = 1.67 × 10-27 kg)
To find the kinetic energy of a relativistic proton, we need to use the equation for relativistic kinetic energy:
K = (γ - 1) * mc^2
where K is the kinetic energy, γ is the Lorentz factor, m is the rest mass, and c is the speed of light.
First, let's calculate the Lorentz factor γ:
γ = 1 / sqrt(1 - (v^2 / c^2))
where v is the velocity of the proton.
Given:
Momentum (p) = 1.0 × 10^-17 kg ∙ m/s
Rest energy (E_0) = 0.15 nJ = 0.15 × 10^-9 J
Mass of the proton (m) = 1.67 × 10^-27 kg
Speed of light (c) = 3.00 × 10^8 m/s
We can find the velocity (v) of the proton using the momentum formula:
p = mv
Solving for v, we get:
v = p / m
Substituting the given values:
v = (1.0 × 10^-17 kg ∙ m/s) / (1.67 × 10^-27 kg)
Calculating v, we get:
v ≈ 5.988 × 10^9 m/s
Now, we can calculate the Lorentz factor γ:
γ = 1 / sqrt(1 - (v^2 / c^2))
Substituting the values:
γ = 1 / sqrt(1 - ((5.988 × 10^9 m/s)^2 / (3.00 × 10^8 m/s)^2))
Calculating γ, we get:
γ ≈ 109.485
Finally, we can calculate the kinetic energy (K):
K = (γ - 1) * mc^2
Substituting the values:
K = (109.485 - 1) * (1.67 × 10^-27 kg) * (3.00 × 10^8 m/s)^2
Calculating K, we get:
K ≈ 4.77 × 10^-11 J
Therefore, the kinetic energy of the proton is approximately 4.77 × 10^-11 Joules.