A beam of alpha particles is incident on a target of lead. A particular alpha pa

Question

A beam of alpha particles is incident on a target of lead. A particular alpha particle comes in "head-on" to a particular lead nucleus and stops 6.10×1014 m away from the center of the nucleus. (This point is well outside the nucleus). Assume that the lead nucleus, which has 82 protons, remains at rest. The mass of the alpha particle is 6.64×1027kg.

A) Calculate the electrostatic potential energy at the instant that the alpha particle stops. Express your result in joules.

U = ____ J

B) Calculate the electrostatic potential energy at the instant that the alpha particle stops. Express your result in MeV.

U = ____ MeV

C) What initial kinetic energy (in joules) did the alpha particle have?

K1 = ____ J

D) What initial kinetic energy (in MeV) did the alpha particle have?

K1 = ____ MeV

E) What was the initial speed of the alpha particle?

V1 = ____ m/s

Explanation / Answer

let

q1 = 2*e (charge of alfa particle)

q2 = 82*e

d = 6.1*10^-14 m

A) U = k*q1*q2/d

= k*2*e*82*e/d

= 164*k*e^2/d

= 164*9*10^9*(1.6*10^-19)^2/(6.1*10^-14)

= 6.19*10^-13 J

B) U = 6.19*10^-13/(1.6*10^-19) eV

= 3.87*10^6 eV

= 3.87 Mev

C) U = 6.19*10^-13 J (using conservation of energy)

D) K1 = 3.87 Mev

E) KE = 0.5*m*v^2

==> v = sqrt(2*KE/m)

= sqrt(2*6.19*10^-13/(6.64*10^-27))

= 1.37*10^7 m/s

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