What is the kinetic energy in MeV of a π–meson that lives 1.80×10-16 s as measured in the laboratory, and 0.840×10-16 s when at rest relative to an observer, given that its rest energy is 135 MeV?
To find the kinetic energy of a π–meson, we need to use the equation for relativistic kinetic energy:
K.E. = (γ - 1) * m_0 * c^2
where γ is the Lorentz factor, m_0 is the rest mass of the particle, and c is the speed of light.
The Lorentz factor, γ, can be calculated using the formula:
γ = 1 / sqrt(1 - (v^2/c^2))
where v is the velocity of the particle.
First, let's find the velocity of the π-meson in the laboratory frame of reference:
Given that the π-meson lives for 1.80×10^-16 s in the laboratory frame, we can use the equation:
Δt_lab = γ * Δt_rest
where Δt_rest is the time measured in the rest frame.
We are given Δt_lab = 1.80×10^-16 s and Δt_rest = 0.840×10^-16 s.
Simplifying the equation, we get:
1.80×10^-16 s = γ * 0.840×10^-16 s
Dividing both sides by 0.840×10^-16 s, we get:
γ = 1.80×10^-16 s / 0.840×10^-16 s
γ = 2.14
Now, let's calculate the velocity (v) using the Lorentz factor formula:
2.14 = 1 / sqrt(1 - (v^2/c^2))
Squaring both sides, we get:
4.5796 = 1 / (1 - (v^2/c^2))
Rearranging the equation, we get:
v^2/c^2 = 1 - 1/4.5796
v^2/c^2 = 0.781
Taking the square root of both sides, we get:
v/c = 0.883
Now, let's calculate the velocity (v) and the Lorentz factor (γ) when the π-meson is at rest relative to an observer:
Δt_lab = γ * Δt_rest
We are given Δt_lab = 0.840×10^-16 s and Δt_rest = 0.840×10^-16 s.
Simplifying the equation, we get:
0.840×10^-16 s = γ * 0.840×10^-16 s
Dividing both sides by 0.840×10^-16 s, we get:
γ = 0.840×10^-16 s / 0.840×10^-16 s
γ = 1
Since the particle is at rest relative to the observer, the velocity (v) is 0. Therefore, we can calculate the kinetic energy using the equation:
K.E. = (γ - 1) * m_0 * c^2
Plugging in the values:
K.E. = (2.14 - 1) * 135 MeV
K.E. = 1.14 * 135 MeV
K.E. = 153.9 MeV
Therefore, the kinetic energy of the π-meson is 153.9 MeV.