An alpha particle (mass 6.64 × 1027 kg, Z =2)approaches to within 1.01 × 1014 m

Question

An alpha particle (mass 6.64 × 1027 kg, Z =2)approaches to within 1.01 × 1014 m of a carbonnucleus (Z = 6). (A) Find the maximum Coulomb force on the alpha particle.Answer in units of N. (B) Find the acceleration of alpha particle at this point. Answer in units of m/s2. (C) Find the potential energy of alpha particle at this point. Answer in units of MeV.
An alpha particle (mass 6.64 × 1027 kg, Z =2)approaches to within 1.01 × 1014 m of a carbonnucleus (Z = 6). (A) Find the maximum Coulomb force on the alpha particle.Answer in units of N. (B) Find the acceleration of alpha particle at this point. Answer in units of m/s2. (C) Find the potential energy of alpha particle at this point. Answer in units of MeV.

Explanation / Answer

a) F = kqaqc/r2 =k(2e)(6e)/r2 = 12ke2/r2 , where kis Coloumb's constant, r is the distance between the particles, ande is the magnitude of the charge of a proton. b) F = maa a = F/ma , where F is the force in part a, andma is the mass of the alpha particle. c) V = -W = -Fr by definition, where F is the force in part a and r isthe distance between the particles.

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