The pion (m_p = 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 DE*Dt = h/2p? (h/2p = 1.05 x 10^-34J*s,c = 3.00 x 10⁸ m/s,and 1 eV = 1.6 x 10^-19 J) A)1.2 x 10^-15 m B)1.4 x 10^-15 m C)2.0 x 10^-15 m D)7.5 x 10^-15 m

Question

Quizplus · Accepted Answer

The maximum range of the particle can be calculated using the formula:

Range = velocity x lifetime

Lifetime can be determined from the uncertainty principle:

Δt = h/2ΔE

where ΔE is the uncertainty in energy. Since the pion mass has an uncertainty of approximately 1 MeV, we can take ΔE as 1 MeV.

Δt = (1.05 x 10^-34)/(2π x 1 MeV)

Δt = 3.33 x 10^-26 s

Velocity of pion is given as 0.99c (where c is the speed of light).

Range = (0.99c) x (3.33 x 10^-26 s)

Range = 1.4 x 10^-15 m

Therefore, the maximum range of the pion is 1.4 x 10^-15 m, which is best represented by option B.

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

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