Uranium-238 is radioactive. A uranium nucleus (mass 238 u, charge 92 e) at rest

Question

Uranium-238 is radioactive. A uranium nucleus (mass 238 u, charge 92 e) at rest decays by emitting an alpha particle (mass 4 u, charge 2 e; essentially a helium nucleus), leaving behind a smaller “daughter” nucleus. This process can by modeled in the following way: the original uranium nucleus is a 7.4 fm radius uniformly charged sphere. Two protons and two neutrons “pinch off” to form the alpha particle, which is initially at rest and immediately adjacent to the remaining nucleus (the daughter). You can assume that at this moment their (center-to-center) separation is 7.4 fm. These two positively charged particles will then repel, “emitting” the alpha.1 (a) What is the mass (in u) and charge (in e) of the daughter nucleus? What is the name of this chemical element? (b) What is the total electrostatic potential energy of the alpha and daughter right after forming? Write your answer in J and in eV. (c) What is the electric field strength of the daughter nucleus at the location of the alpha? What is force experience by the alpha at this location? What is its initial acceleration due to this force? (d) Find the speed of the daughter and the alpha after they have moved far apart. Write your answer in m/s and as a fraction of c, the speed of light. [Hint: this “collision” conserves momentum and mechanical energy. The “before” state is the daughter and alpha next to each other at rest and the “after” state is the two moving at infinite separation.]

Explanation / Answer

1. U-238 looses 4u mass and 2e charge after emiting the alpha particel

              mass of daughter = 238 -4 = 234 u

                            charge   = 92 - 2 = 90e

daughter nucleus is Th-234 , Thorium

charge on daughter nucleus = 90e

charge on alpha = 2e

distance of seperation d = 7.4 fm

Potential energy U = kq1q2/d = 9.0E+9 *90*2e2 /7.4E-15

                                 = 5.6 E-12 J   ( e = 1.6e-19 C)

                                  = 35 Mev      ( 1 ev = 1.6 e-19 J)

c) charge on the daughter = 90e

     distance r = 7.4 E-15 m

     electric field at the location of alpha E = kq/r2 = 9.0E+9 *90*1.6E-19/(7.4E-15)2

                                                                                = 23.67 E+20 N/C

Force on the aplha particle F = Eq = 23.67E+20 *2*1.6E-19 = 757.44 N

mass of aplha m= 4u = 4* 1.66E-27 kg

initial acceleration   a= F/m = 757.44/(4*1.66E-27) = 114.07 E+27 m/s2

d) When the daughter and the alpha are seperated far apart the PE of the system is 0. Conserving mechnical energy initial PE of the system is equal to the KE of the alpha, daughter is at rest.

KE of the aplha = 5.6 E-12 J

             speed v = sqrt( 2*5.6E-12/4*1.66E-27) = 4.11E+7 m/s

                            = 4.11E+7/3.0E+8 = 0.137c

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