(a) Define
(i) proton number;
(ii) nucleon number;
(iii) isotopes.
(b) A nuclide \(^A_ZX\) emits \(\beta\)-particle to form a daughter nuclide Y. Write a nuclear equation to illustrate the charge conservation.
(c) The radioactive nuclei \(^{210}_{84}P_o\) emits an \(\alpha\) - particle to produce \(^{206}_{82}P_b\). Calculate the energy, in MeV, released in each disintegration.
Take the masses of \(^{210}_{84}P_o\) = 209.936730 u;
\(^{206}_{82}P_b\) = 205.929421 u;
\(^{4}_{2}He\) = 4.001504 u;
and that 1u = 931 MeV
(a)(i) Proton number (atomic number): the number of protons in the nucleus of an atom.
(ii) Nucleon number (mass number): the total number of protons and neutrons in the nucleus of an atom.
(iii) Isotopes: atoms of the same element that have the same proton number but different nucleon numbers (same number of protons, different numbers of neutrons).
(b) Beta emission converts a neutron into a proton, so the proton number rises by 1 while the nucleon number is unchanged:
\[ {}^{A}_{Z}X \rightarrow {}^{A}_{Z+1}Y + {}^{0}_{-1}e \]
Charge conservation: \(Z = (Z+1) + (-1)\), which balances.
(c) Mass defect in the disintegration:
\[ \Delta m = m({}^{210}_{84}Po) - \left[ m({}^{206}_{82}Pb) + m({}^{4}_{2}He) \right] \]
\[ \Delta m = 209.936730 - (205.929421 + 4.001504) \]
\[ \Delta m = 209.936730 - 209.930925 = 0.005805\ \text{u} \]
Energy released:
\[ E = \Delta m \times 931 = 0.005805 \times 931 \]
\[ E = 5.40\ \text{MeV} \]
The energy released in each disintegration is about \(5.40\ \text{MeV}\).
(a)(i) Proton number (atomic number): the number of protons in the nucleus of an atom.
(ii) Nucleon number (mass number): the total number of protons and neutrons in the nucleus of an atom.
(iii) Isotopes: atoms of the same element that have the same proton number but different nucleon numbers (same number of protons, different numbers of neutrons).
(b) Beta emission converts a neutron into a proton, so the proton number rises by 1 while the nucleon number is unchanged:
\[ {}^{A}_{Z}X \rightarrow {}^{A}_{Z+1}Y + {}^{0}_{-1}e \]
Charge conservation: \(Z = (Z+1) + (-1)\), which balances.
(c) Mass defect in the disintegration:
\[ \Delta m = m({}^{210}_{84}Po) - \left[ m({}^{206}_{82}Pb) + m({}^{4}_{2}He) \right] \]
\[ \Delta m = 209.936730 - (205.929421 + 4.001504) \]
\[ \Delta m = 209.936730 - 209.930925 = 0.005805\ \text{u} \]
Energy released:
\[ E = \Delta m \times 931 = 0.005805 \times 931 \]
\[ E = 5.40\ \text{MeV} \]
The energy released in each disintegration is about \(5.40\ \text{MeV}\).