In the lab, a relativistic proton has a momentum of 1.00 ×10^-19 kg ∙ m/s and a rest energy of 0.150 nJ. What is the speed
of the proton in the lab? (c = 3.00 × 10^8 m/s, mass of proton = 1.67 ×10^-27 kg)
0,170c
To find the speed of the proton in the lab, we can use the following equation:
E = γmc^2
where E is the total energy of the proton, γ is the Lorentz factor, m is the mass of the proton, and c is the speed of light.
The total energy of the proton can be calculated as the sum of its rest energy and kinetic energy:
E = mc^2 + K
where K is the kinetic energy of the proton.
Now, we can express the kinetic energy of the proton in terms of its momentum:
K = √(p^2c^2 + m^2c^4) - mc^2
where p is the momentum of the proton.
Substituting the values given in the problem:
p = 1.00 × 10^-19 kg ∙ m/s
m = 1.67 × 10^-27 kg
c = 3.00 × 10^8 m/s
we can now solve for the kinetic energy:
K = √((1.00 × 10^-19 kg ∙ m/s)^2 * (3.00 × 10^8 m/s)^2 + (1.67 × 10^-27 kg)^2 * (3.00 × 10^8 m/s)^4) - (1.67 × 10^-27 kg) * (3.00 × 10^8 m/s)^2
Once we have the kinetic energy, we can substitute it back into the equation for total energy to solve for γ:
E = γmc^2
Plugging in the values:
0.150 nJ = γ * (1.67 × 10^-27 kg) * (3.00 × 10^8 m/s)^2 + K
we can solve for γ:
γ = (0.150 nJ - K) / ((1.67 × 10^-27 kg) * (3.00 × 10^8 m/s)^2)
Finally, we can find the speed of the proton in the lab by using the relationship between γ and v:
γ = 1 / √(1 - v^2 / c^2)
Rearranging the above equation, we have:
v = √(1 - (1 / γ)^2) * c
Substituting the value of γ obtained earlier, we can calculate the speed of the proton in the lab:
v = √(1 - (1 / ((0.150 nJ - K) / ((1.67 × 10^-27 kg) * (3.00 × 10^8 m/s)^2)))^2) * (3.00 × 10^8 m/s)