QUESTION:
The isotope 218Po decays via α−decay. The measured atomic mass of 218Po is 218.00897 u, and the atomic mass of the daughter nucleus is 213.99981 u. Find
the Name of the daughter nucleus.
the number of nucleons in the daughter nucleus.
the atomic number of the daughter nucleus.
the number of neutrons in the daughter nucleus.
the kinetic energy of the α−particle. (Ignore the recoil of the daughter nucleus.)
SOLUTION:
Given:
- The isotope 218Po decays via α−decay.
- The atomic mass of the parent nucleus, 218Po is 218.00897 u, and
- The atomic mass of the daughter nucleus is 213.99981 u.
We know the values:
- The atomic number of 218Po is 84.
- The atomic number of Pb214 is 82.
- The unified atomic mass, u=931.5 MeV/c2
As the α−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−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 u which is four less than the mass number of the parent nucleus, 218Po. It means only one α−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 214Pb.
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 218Po+214Pb→ 4He is only in the form of the kinetic energy of the α−particle then the energy equivalent to the mass defect will be equal to the kinetic energy of the α−particle. In another word, we can say that the kinetic energy of the α−particle is provided by the energy equivalent to the mass defect (Δmc2). Here Δm is the mass defect and c is the speed of light in a vacuum.
The mass defect is given by;
Δm=Mass(218Po)–Mass(214Pb)−Mass(4He)=218.00897 u–213.99981 u–4.002602 u)=0.006558 u=0.006558×931.5 MeV/c2=6.108777 MeV/c2
Therefore, the kinetic energy of the α−particle is;
K.E.α=Δmc2=(6.108777 MeV/c2)×c2=6.108777 MeV
Similar Problems based on Mass-Energy Equivalence
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A free neutron can decay into a proton, an electron, and an anti-neutrino. Assume the anti-neutrino’s rest mass is zero, and the rest masses for proton and electron are 1.6726×10−27 kg and 9.11×10−31 kg respectively. Determine the total kinetic energy shared among the three particles when a neutron decays at rest.
The unified atomic mass unit, denoted by u, is defined to be 1u=1.6605×10−27 kg. It can be used as an approximation for the average mass of a nucleon in a nucleus, taking the binding energy into account. Find the energy obtained after converting a nucleus of 14 nucleons completely into free energy.
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