Calculate the range of the force that might be produced by the virtual exchange of a proton.Assume ΔE⋅Δt = h/2π.(m_p = 1.67 × 10−²⁷ kg,c = 3.00 × 10⁸ m/s,and h/2π = 1.05 × 10−³⁴ J⋅s) A)6.72 × 10−²⁵ m B)2.10 × 10−¹⁶ m C)6.00 × 10−⁸ m D)1.54 × 10−¹⁵ m

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The range of the force can be estimated using the uncertainty principle, ΔE⋅Δt ≈ h/2π, where ΔE is the energy associated with the virtual exchange and Δt is the time for which this energy is available. The energy can be related to the mass of the proton (m_p) and the speed of light (c) using the relation ΔE = m_p*c^2. Substituting the given values, ΔE = 1.67 × 10^(-27) kg * (3.00 × 10^8 m/s)^2 = 1.50 × 10^(-10) J. Using ΔE⋅Δt = h/2π = 1.05 × 10^(-34) J⋅s, we find Δt = 1.05 × 10^(-34) J⋅s / 1.50 × 10^(-10) J = 7.00 × 10^(-25) s. The range, which can be approximated by the distance light travels in time Δt, is c*Δt = 3.00 × 10^8 m/s * 7.00 × 10^(-25) s = 2.10 × 10^(-16) m.

Calculate the Range of the Force That Might Be Produced