Question

Plutonium-239 is one of the major isotopes used in nuclear reactors and nuclear weapons.
It has a half-life of 24,100 years.

a.) how many alpha and beta decays are needed for 239/94PU to decay 227/89Ac?

b.) Calculate the kinetic energy released from the alpha decay of s39/94Pu. Compare this to the
kinetic energy released by the fission reaction.

n + 239/94 Pu-->100/40 Zr + 137/54 Xe + 3n

Answer 1

[a]

PU²³⁹₉₄ decays to Ac ₈₉²²⁷

Here mass number reduces to 239-227= 12

and atomic number reduces to 94-89= 5

Hence three particles and one particle is emitted.

[b]

kinetic energy released from the alpha decay of Pu²³⁹₉₄ is

PU²³⁹₉₄ ---------------------- He⁴₂+X²³⁵₉₂

Apply

Q= [KE_after?Ke_before ]Q = (Rest mass_before-Rest mass_after )c²

Q= {2.46 u- [0.033u+1.96u]} c²=0.467u*c²                    [1u=1.67*10^-27 kg]

Q=0.467*1.67*10^-27 *[3*10⁸ ]² =7.02*10^-11J

kinetic energy released by the fission reaction. n¹₀ + Pu²³⁹₉₄-->Zr¹⁰⁰₄₀ + Xe¹³⁵₅₄+ 3n¹₀

Q= (Rest mass_before-Rest mass_after )c²

Q = [{2.46u+1.00867u}-{0.834u+1.13u} ] c² = [ 3.469u-1.964u]c² = 1.505u*c²

Q=1.505u*c² =1.505*1.67*10^-27 *[3*10⁸ ]² =2.26*10^-10J                          [1u=1.67*10^-27 kg]

Hence kinetic energy released by the fission reaction greater than kinetic energy released from the alpha decay of Pu²³⁹₉₄