An alpha particle ( Z = 2, mass 6.64 x 10 -27 kg)approaches to within 1.00 x10 -

Question

An alpha particle (Z = 2, mass 6.64 x 10-27 kg)approaches to within 1.00 x10-14 m of a carbon nucleus (Z =6). (a) What is the maximum Coulomb force on thealpha particle?
1 N
(b) What is the acceleration of the alpha particle at thispoint?
2 m/s2
(c) What is the potential energy of the alpha particle at the sametime?
3 MeV (a) What is the maximum Coulomb force on thealpha particle?
1 N
(b) What is the acceleration of the alpha particle at thispoint?
2 m/s2
(c) What is the potential energy of the alpha particle at the sametime?
3 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 x10-14 m]2 = 27.61 N b) acceleration = Force/mass = 27.61 N / [6.64 x 10-27 kg]= = 4.15809589 x 1027 m/s2 ~ 4.16 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 x 10-14 m] Potential energy = 2.76 x 10-13 N*m = 2.76 x10-13 J 2.76 x 10-13 J * [1eV / 1.6 x 10-19 J]* [1MeV / 1 x106 eV] = 1.73 MeV

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