(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
(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
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Correct Answer: Option n
Explanation:
(a)(i) Proton number
— the number of protons in the nucleus of an atom
(ii) Nucleon number — the total number of protons and neutrons in the nucleus of are atom.
(iii) Isotopes — are atoms or nuclides of the samc element having the same number of protons or atomic number but different neutrons or mass numbers
(b) \(^A_Z \to ^A_{Z+}Y + ^0_{-1}\)
\(^{210}_{84}Po \to ^{206}_{82}Pb + ^4_2He + Energy\)
Mass detect = 209.936730u - (205.929421 + 4.001504)u
209.9367330u - 209.930925u
= 5.805 x 10\(^{-3}\)U
Energy released = 5.805 x 10\(^{-3}\) x 931
= 5.40MeV
(a)(i) Proton number
— the number of protons in the nucleus of an atom
(ii) Nucleon number — the total number of protons and neutrons in the nucleus of are atom.
(iii) Isotopes — are atoms or nuclides of the samc element having the same number of protons or atomic number but different neutrons or mass numbers
(b) \(^A_Z \to ^A_{Z+}Y + ^0_{-1}\)
\(^{210}_{84}Po \to ^{206}_{82}Pb + ^4_2He + Energy\)
Mass detect = 209.936730u - (205.929421 + 4.001504)u
209.9367330u - 209.930925u
= 5.805 x 10\(^{-3}\)U
Energy released = 5.805 x 10\(^{-3}\) x 931
= 5.40MeV