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Question 11 Mark
Show that the minimum energy needed to separate a proton from a nucleus with $Z$ protons and $N$ neutrons is:
$\Delta\text{E}=(\text{M}_{\text{Z}-1,\text{N}}+\text{M}_{\text{H}}-\text{M}_{\text{Z,N}})\text{c}^2$
where $M_{Z,N} =$ mass of an atom with $Z$ protons and $N$ neutrons in the nucleus and $M_H =$ mass of a hydrogen atom. This energy is known as proton-separation energy.
Answer
$E_{Z.N.} \rightarrow E_{Z-1,} N + P_1$
$\Rightarrow E_{Z.N.} \rightarrow E_{Z-1,} N + _1H^1 [$As hydrogen has no neutrons but protons only$]$
$\Delta E = (M_{Z-1, }N + N_H - M_{Z,N})c^2$
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Question 21 Mark
  1. Calculate the energy released if $^{238}U$ emits an $\alpha-$particle.
  2. Calculate the energy to be supplied to $^{238}U$ it two protons and two neutrons are to be emitted one by one. The atomic masses of $^{238}U,\ ^{234}Th$ and $^4He$ are $238.0508u, 234.04363u$ and $4.00260u$ respectively.
Answer
  1. $U^{238}\ _2He^4 + Th^{234}$
$E = [Mu - (N_{HC} + M_{Th})]u = 238.0508 - (234.04363 + 4.00260)]u = 4.25487Mev = 4.255Mev.$
  1. $E = U^{238} - [Th^{234} + 2n'_0 + 2p'_1]$
$= {238.0508 - [234.64363 + 2(1.008665) + 2(1.007276)]}u$
$= 0.024712u = 23.0068 = 23.007MeV.$
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Question 31 Mark
$^{32}P$ beta-decays to $^{32}S.$ Find the sum of the energy of the antineutrino and the kinetic energy of the $\beta-$particle. Neglect the recoil of the daughter nucleus. Atomic mass of $^{32}P = 31.974u$ and that of $^{32}S = 31.972u.$
Answer
$\text{P}^{32}\rightarrow\text{S}^{32}+ \ _0\bar{\text{v}}^0+ \ _1\beta^0$
Energy of antineutrino and $\beta$-particle
$= (31.974 - 31.972)u = 0.002u = 0.002 \times 931 = 1.862\ MeV = 1.86.$
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Question 41 Mark
Calculate the minimum energy needed to separate a neutron from a nucleus with $Z$ protons and $N$ neutrons it terms of the masses $M_{Z.N,} M_{Z,N-1}$ and the mass of the neutron.
Answer
$\text{E}_2\text{N}=\text{E}_{\text{Z,N}-1}+\text{ }^1_0\text{n}.$
Energy released $= ($Initial Mass of nucleus $-$ Final mass of nucleus$)c^2 $
$= (M_{Z.N-1} + M_0 - M_{ZN})c^2.$
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