An alpha particle (mass 6.64e-27 kg, Z = 2) approches towithin 1.28e-14 m of a c

Question

An alpha particle (mass 6.64e-27 kg, Z = 2) approches towithin 1.28e-14 m of a carbon nucleus (Z=6).
a) Find the maximum Coulomb force on the alpha particle. Answer inunits 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 6.64e-27 kg, Z = 2) approches towithin 1.28e-14 m of a carbon nucleus (Z=6).
a) Find the maximum Coulomb force on the alpha particle. Answer inunits 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)   Force =k*q1*q2/r2 k = 8.987551 x 109 Nm2/C2 q1 = Z*e = 2*1.6 x 10-19C        (e = unit charge = 1.6x 10-19 C) q2 = Z*e = 6*1.6 x 10-19 C Force = [8.987551 x 109 Nm2/C2] *[2*1.6 x 10-19 C] [6*1.6 x 10-19 C] / [1.28 x10-14 m]2 = 16.85 N b) acceleration = Force/mass = 16.85 N / [6.64 x 10-27 kg]= 2.53408392 x 1027 m/s2 ~ 2.53 x1027 m/s2 c) Potential energy = k*q1*q2/r Potential energy = [8.987551 x 109Nm2/C2] * [2*1.6 x 10-19 C] [6*1.6x 10-19 C] / [1.28 x 10-14 m] Potential energy = 2.16 x 10-13 N*m = 2.16 x10-13 J 2.16 x 10-13 J * [1eV / 1.6 x 10-19 J]* [1MeV / 1 x106 eV] = 1.35 MeV

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