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Physics 1 Marks Question

The Nucleus · Jharkhand Board (JAC) STD 12 Science (Jharkhand - English Medium). 4 questions.

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