Conceptual Question on Mass-Energy Equivalence. #2

Given:

• The isotope \rm ^{218}Po decays via \alpha-decay.
• The atomic mass of the parent nucleus, \rm ^{218}Po is 218.00897\rm \ u, and
• The atomic mass of the daughter nucleus is 213.99981\rm \ u.

We know the values:

• The atomic number of \rm ^{218}Po is 84.
• The atomic number of \rm Pb^{214} is 82.
• The unified atomic mass, \rm u = 931.5 \ MeV/c^2

As the \alpha-particle is structurally equivalent to the nucleus of a helium atom (consists of two protons and two neutrons) so after the decay of an \alpha-particle, the mass number of the parent nucleus is decreased by four and the atomic number is decreased by two.

It is given that the mass number of the daughter nucleus is 213.99981\rm \ u which is four less than the mass number of the parent nucleus, \rm ^{218}Po. It means only one \alpha-particle is emitted. So, the atomic number of the daughter nucleus is two less than the atomic number of the parent nucleus. It is 82.

Part (1)

The daughter nucleus is \rm ^{214}Pb.

Part (2)

The nucleon number of the daughter nucleus is N = 214.

Part (3)

The atomic number of the daughter nucleus is Z = 82.

Part (4)

The neutron number of the daughter nucleus is n=214 -82 = 132.

Part (5)

If suppose, the energy released in the reaction \rm ^{218}Po+^{214}Pb \rightarrow \ ^4He is only in the form of the kinetic energy of the \alpha-particle then the energy equivalent to the mass defect will be equal to the kinetic energy of the \alpha-particle. In another word, we can say that the kinetic energy of the \alpha-particle is provided by the energy equivalent to the mass defect (\Delta m c^2). Here \Delta m is the mass defect and c is the speed of light in a vacuum.

The mass defect is given by;

\begin{aligned} \Delta m &= \text{Mass}(\rm ^{218}Po) – \text{Mass} (\rm ^{214}Pb)- \text{Mass} (^{4}He)\\ &= 218.00897\rm \ u – 213.99981\rm \ u – 4.002602\rm \ u)\\ &= 0.006558 \rm \ u \\ &= 0.006558 \times 931.5\rm \ MeV/c^2\\ &= 6.108777\rm \ MeV/c^2\\ \end{aligned}

Therefore, the kinetic energy of the \alpha-particle is;

\begin{aligned} K.E._{\alpha} &= \Delta m c^2\\ &=(6.108777 \ MeV/c^2) \times c^2\\ &= 6.108777 \ MeV \\ \end{aligned}