Final answer:
The Lorentz factor γ for a proton accelerated through 1.0 TV at Fermilab, with a mass energy of 938.3 MeV, is calculated to be approximately 1065. The closest given option to this value is (c) (1.0 × 10^9).
Step-by-step explanation:
The question asks to compute the Lorentz factor, γ (gamma), for a proton that is accelerated through an effective potential of 1.0 teravolt (TV) at Fermilab. The Lorentz factor is given by γ = E / (mc^2), where E is the total energy of the proton and mc^2 is the mass energy of the proton.
m is the mass of the proton and c is the speed of light. The mass energy of a proton is given as 938.3 MeV, and since the proton is accelerated through a potential of 1.0 TV, its total energy E will be the mass energy plus the energy acquired from acceleration, which in this case equals to 1 TV = 1.0 × 10^12 eV since 1eV = 1.602 × 10^-19 Joules.
To find γ for the proton, first convert the mass energy of the proton to electron volts (eV): 938.3 MeV = 938.3 × 10^6 eV. Then add the kinetic energy obtained from the acceleration to get the total energy E. Assuming that the proton gains energy equal to the accelerating potential, E = mass energy + kinetic energy = 938.3 × 10^6 eV + 1.0 × 10^12 eV = 1.0 × 10^12 eV + 938.3 × 10^6 eV = approximately 1.0 × 10^12 eV, as the additional mass energy is negligible compared to the kinetic energy gained.
We can now calculate γ as follows:
γ = E / (mc^2) = (1.0 × 10^12 eV) / (938.3 × 10^6 eV) ≈ 1065
Therefore, the correct answer closest to our calculation is option (c) (1.0 × 10^9).