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2006 WAEC Physics Theory (a) Define (i) proton number; (ii) nucleon number; (iii) isotopes. (b) A nuclide \(^A_ZX\) emits...

Physics
WAEC 2006

(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

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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


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