Conceptual Question on Mass-Energy Equivalence. #2

 QUESTION:

The isotope 218Po \rm ^{218}Po decays via α \alpha-decay. The measured atomic mass of 218Po \rm ^{218}Po is 218.00897 u218.00897\rm \ u, and the atomic mass of the daughter nucleus is 213.99981 u213.99981\rm \ u. Find
  1. the Name of the daughter nucleus.
  2. the number of nucleons in the daughter nucleus.
  3. the atomic number of the daughter nucleus.
  4. the number of neutrons in the daughter nucleus.
  5. the kinetic energy of the α \alpha-particle. (Ignore the recoil of the daughter nucleus.)

 SOLUTION:

Given:

  • The isotope 218Po \rm ^{218}Po decays via α\alpha-decay.
  • The atomic mass of the parent nucleus, 218Po \rm ^{218}Po is 218.00897 u218.00897\rm \ u, and
  • The atomic mass of the daughter nucleus is 213.99981 u213.99981\rm \ u.

We know the values:

  • The atomic number of 218Po\rm ^{218}Po is 84.
  • The atomic number of Pb214\rm Pb^{214} is 82.
  • The unified atomic mass, u=931.5 MeV/c2\rm u = 931.5 \ MeV/c^2

As the α\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 α\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 u213.99981\rm \ u which is four less than the mass number of the parent nucleus, 218Po\rm ^{218}Po. It means only one α\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 8282.

Part (1)

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

Part (2)

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

Part (3)

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

Part (4)

The neutron number of the daughter nucleus is n=21482=132n=214 -82 = 132.

Part (5)

If suppose, the energy released in the reaction 218Po+214Pb 4He\rm ^{218}Po+^{214}Pb \rightarrow \ ^4He is only in the form of the kinetic energy of the α\alpha-particle then the energy equivalent to the mass defect will be equal to the kinetic energy of the α\alpha-particle. In another word, we can say that the kinetic energy of the α\alpha-particle is provided by the energy equivalent to the mass defect (Δmc2\Delta m c^2). Here Δm\Delta m is the mass defect and cc is the speed of light in a vacuum.

The mass defect is given by;

Δm=Mass(218Po)–Mass(214Pb)Mass(4He)=218.00897 u213.99981 u4.002602 u)=0.006558 u=0.006558×931.5 MeV/c2=6.108777 MeV/c2\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 α\alpha-particle is;

K.E.α=Δmc2=(6.108777 MeV/c2)×c2=6.108777 MeV\begin{aligned} K.E._{\alpha} &= \Delta m c^2\\ &=(6.108777 \ MeV/c^2) \times c^2\\ &= 6.108777 \ MeV \\ \end{aligned}

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