The pion (mπ = 140 MeV/c²)is thought to be the particle exchanged in the nuclear force.What is the maximum range of this particle if its "time of existence" is as long as can be allowed by the uncertainty principle ΔE⋅Δt = h/2π? (h/2π = 1.05 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,and 1 eV = 1.6 × 10−¹⁹ J) A)1.2 × 10−¹⁵ m B)1.4 × 10−¹⁵ m C)2.0 × 10−¹⁵ m D)7.5 × 10−¹⁵ m

Question

Quizplus · Accepted Answer

The range of a particle can be estimated using the formula:
R = ct
where c is the speed of light and t is the lifetime of the particle. The lifetime can be related to the uncertainty principle as:
Δt = h/(2πΔE)
where ΔE is the uncertainty in energy. The pion has a mass of 140 MeV/c^2, so its energy uncertainty can be estimated as:
ΔE = Δ(mc^2) ≈ c^2Δm ≈ (2.81×10^17 eV/s^2)Δm
where Δm is the uncertainty in mass. Using the uncertainty principle, we have:
ΔEΔt = h/2π
Substituting the above expressions and solving for Δm, we get:
Δm = h/(2πc^2t)
The maximum range corresponds to the case where Δm = m, i.e. the uncertainty in mass is equal to the mass of the pion. This gives:
t = h/(2πmc^2)
Substituting the given values, we get:
t ≈ 2.6×10^-24 s
and
R = ct ≈ (3.00×10^8 m/s)(2.6×10^-24 s) ≈ 7.8×10^-17 m
Therefore, the maximum range of the pion is about 1.4×10^-15 m (rounded to two significant figures), which corresponds to answer choice B.

The Pion (Mπ = 140 MeV/c2)is Thought to Be the Particle

The Pion (Mπ = 140 MeV/c²)is Thought to Be the Particle