46: The Nucleus : Neet-xii-physics

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NEET-XII-Physics

46: The Nucleus

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Qstn #9
Two lithium nuclei in a lithium vapour at room temperature do not combine to form a carbon nucleus because
(a) a lithium nucleus is more tightly bound than a carbon nucleus
(b) carbon nucleus is an unstable particle
(c) it is not energetically favourable
(d) Coulomb repulsion does not allow the nuclei to come very close.
digAnsr: d
Ans : (d) Coulomb repulsion does not allow the nuclei to come very close.
Lithium atom contains 3 protons and 3 neutrons in the nucleus and 3 valence electrons. When two lithium nuclei are brought together, they repel each other. The attractive nuclear forces being short-range are insignificant as compared to the electrostatic repulsion. Thus, the nuclei do not combine to form carbon atom because of coulomb repulsion.
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Qstn #10
For nuclei with A > 100,
(a) the binding energy of the nucleus decreases on an average as A increases
(b) the binding energy per nucleon decreases on an average as A increases
(c) if the nucleus breaks into two roughly equal parts, energy is released
(d) if two nuclei fuse to form a bigger nucleus, energy is released.
digAnsr: b,c,A
Ans : (b) the binding energy per nucleon decreases on an average as A increases
(c) if the nucleus breaks into two roughly equal parts, energy is released
Binding energy per nucleon varies in a way that it depends on the actual value of mass number (A). As the mass number (A) increases, the binding energy also increases and reaches its maximum value of 8.7 MeV for A (50-80) and for A > 100. The binding energy per nucleon decreases as A increases and the nucleus breaks into two or more atoms of roughly equal parts so as to attain stability and binding energy of mass number between 50-80.
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#
Section : iv

Qstn #1
Assume that the mass of a nucleus is approximately given by M = Am_p where A is the mass number. Estimate the density of matter in kgm^-3 inside a nucleus. What is the specific gravity of nuclear matter?
Ans : Given:
Mass of the nucleus, M = Am_p
Volume of the nucleus, V = `` \frac{4}{3}\pi {{R}_{0}}^{3}A``
Density of the matter, `` d=\frac{M}{V}=\frac{A{m}_{p}}{{\displaystyle \frac{4}{3}}\pi {{R}_{0}}^{3}A}``
`` =\frac{3{m}_{p}}{4\times \pi {{R}_{0}}^{3}}``
`` =\frac{3\times 1.007276}{4\times 3.14(1.1{)}^{3}}``
`` =3\times {10}^{17}\,\mathrm{\,kg\,}/{\,\mathrm{\,m\,}}^{3}``
Specific gravity of the nuclear matter = `` \frac{\,\mathrm{\,Density\,}\,\mathrm{\,of\,}\,\mathrm{\,matter\,}}{\,\mathrm{\,Density\,}\,\mathrm{\,of\,}\,\mathrm{\,water\,}}``
`` \therefore `` Specific gravity = `` \frac{3\times {10}^{17}}{{10}^{3}}`` = 3 `` \times `` 10¹⁴ kg/m³
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Qstn #2
A neutron star has a density equal to that of the nuclear matter. Assuming the star to be spherical, find the radius of a neutron star whose mass is 4.0 × 10³⁰ kg (twice the mass of the sun).
Ans : Given:
Mass of the neutron star, M = 4.0 × 10³⁰ kg
Density of nucleus, d = 2.4`` \times ``10¹⁷`` {}^{}``
Density of nucleus, `` d=\frac{M}{V}``
Here, V is the volume of the nucleus.
`` \therefore V=\frac{M}{d}=\frac{4\times {10}^{30}}{2.4\times {10}^{17}}``
`` =\frac{1}{0.6}\times {10}^{13}``
`` =\frac{1}{6}\times {10}^{14}``
`` ``
If R is the radius, then the volume of the neutron star is given by
`` V=\frac{4}{3}\pi {R}^{3}``
`` \therefore \frac{1}{6}\times {10}^{14}=\frac{4}{3}\times \,\mathrm{\,\pi \,}\times {R}^{3}``
`` \Rightarrow {R}^{3}=\frac{1}{6}\times \frac{3}{4}\times \frac{1}{\,\mathrm{\,\pi \,}}\times {10}^{14}``
`` \Rightarrow {R}^{3}=\frac{1}{8}\times \frac{100}{\,\mathrm{\,\pi \,}}\times {10}^{12}``
`` \therefore R=\frac{1}{2}\times {10}^{4}\times 3.17``
`` =1.585\times {10}^{4}=15\,\mathrm{\,km\,}``
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Qstn #3
Calculate the mass of an α-particle. Its Its binding energy is 28.2 MeV.
Ans : Given:
Binding energy of α particle = 28.2 MeV
Let x be the mass of α particle.
We know an α particle consists of 2 protons and 2 neutrons.
Binding energy, `` B=\left(Z{m}_{p}+N{m}_{n}-M\right){c}^{2}``
`` ``
Here, m_p = Mass of proton
m_n = Mass of neutron
Z= Number of protons
N = Number of neutrons
c = Speed of light
On substituting the respective values, we have
`` 28.2=(2\times 1.007276+2\times 1.008665-x){c}^{2}``
`` \Rightarrow x=4.0016\,\mathrm{\,u\,}``
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Qstn #4
How much energy is released in the following reaction:
⁷Li + p → α + α.
Atomic mass of ⁷Li = 7.0160 u and that of ⁴He = 4.0026 u.
Ans : Given:
Mass of ⁷Li = 7.0160 u
Mass of ⁴He = 4.0026 u.
Reaction:
`` \,\mathrm{\,L\,}{i}^{7}+p\mathit{\to }\alpha \mathit{+}\alpha \mathit{+}E\mathit{,}``
`` ``
`` ``
Energy release `` \left(E\right)`` is given by
`` E=\left[m\left({}^{7}\,\mathrm{\,Li\,}\right)+\left({m}_{p}\right)-2\times m\left({}^{4}\,\mathrm{\,He\,}\right)\right]{c}^{2}``
`` =\left[\left(7.0160\,\mathrm{\,u\,}+1.007276\,\mathrm{\,u\,}\right)-2\left(4.0026\,\mathrm{\,u\,}\right)\right]{c}^{2}``
`` =(8.023273\,\mathrm{\,u\,}-8.0052\,\mathrm{\,u\,}){\,\mathrm{\,c\,}}^{2}``
`` =0.018076\times 931\,\mathrm{\,MeV\,}``
`` =16.83\,\mathrm{\,MeV\,}``
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Qstn #5
Find the binding energy per nucleon of
79197Au if its atomic mass is 196.96 u.
Ans : Given:
Atomic mass of Au, A = 196.96
Atomic number of Au, Z = 79
Number of neutrons, N = 118
Binding energy, `` B=(\,\mathrm{\,Z\,}{m}_{p}+N{m}_{n}-M){c}^{2}``
`` ``
`` ``
Here, m_p = Mass of proton
M = Mass of nucleus
m_n = Mass of neutron
c = Speed of light
On substituting the respective values, we get
`` B=\left[\right(79\times 1.007276+118\times 1.008665)\,\mathrm{\,u\,}-196.96\,\mathrm{\,u\,}]{c}^{2}``
`` =\left(198.597274-196.96\right)\times 931\,\mathrm{\,MeV\,}``
`` =1524.302094\,\mathrm{\,MeV\,}``
Binding energy per nucleon `` =\frac{1524.3}{197}=7.737\,\mathrm{\,MeV\,}``
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#6-a
Calculate the energy released if ²³⁸U emits an α-particle.
Ans : Given:
Atomic mass of ²³⁸U, m(²³⁸U) = 238.0508 u
Atomic mass of ²³⁴Th, m(²³⁴Th) = 234.04363 u
Atomic mass of ⁴He, m(⁴He) = 4.00260 u
When ²³⁸U emits an α-particle, the reaction is given by
`` {\,\mathrm{\,U\,}}^{238}\to {\,\mathrm{\,Th\,}}^{234}+{\,\mathrm{\,He\,}}^{4}``
Mass defect, Δm = [m(²³⁸U)`` -``(m(²³⁴Th)+m(⁴He))]
Δm = [238.0508 `` -`` (234.04363 + 4.00260) = 0.00457 u
Energy released (E) when ²³⁸U emits an α-particle is given by
`` E=∆m{c}^{2}``
`` E=[0.00457\,\mathrm{\,u\,}]\times 931.5\,\mathrm{\,MeV\,}``
`` \Rightarrow E=4.25467\,\mathrm{\,MeV\,}=4.255\,\mathrm{\,MeV\,}``
`` ``

#6-b
Calculate the energy to be supplied to ²³⁸U it two protons and two neutrons are to be emitted one by one. The atomic masses of ²³⁸U, ²³⁴Th and ⁴He are 238.0508 u, 234.04363 u and 4.00260 u respectively.
Ans : When two protons and two neutrons are emitted one by one, the reaction will be
`` {\,\mathrm{\,U\,}}^{233}\to {\,\mathrm{\,Th\,}}^{234}+2n+2p``
`` \,\mathrm{\,Mass\,}\,\mathrm{\,defect\,},∆m=m\left({\,\mathrm{\,U\,}}^{238}\right)-[m\left({\,\mathrm{\,Th\,}}^{234}\right)+2\left({m}_{\,\mathrm{\,n\,}}\right)+2\left({m}_{p}\right)]``
`` ∆m=238.0508\,\mathrm{\,u\,}-[234.04363\,\mathrm{\,u\,}+2(1.008665)\,\mathrm{\,u\,}+2(1.007276)\,\mathrm{\,u\,}]``
`` ∆m=0.024712\,\mathrm{\,u\,}``
Energy released (E) when ²³⁸U emits two protons and two neutrons is given by
`` E=∆m{c}^{2}``
`` E=0.024712\times 931.5\,\mathrm{\,MeV\,}``
`` E=23.019=23.02\,\mathrm{\,MeV\,}``
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Qstn #7
Find the energy liberated in the reaction
²²³Ra → ²⁰⁹Pb + ¹⁴C.
The atomic masses needed are as follows.
²²³Ra ²⁰⁹Pb ¹⁴C
22..018 u 208.981 u 14.003 u
Ans : Given:
Atomic mass of ²²³Ra, m(²²³Ra) = 223.018 u
Atomic mass of ²⁰⁹Pb, m(²⁰⁹Pb) = 208.981 u
Atomic mass of ¹⁴C, m(¹⁴C) = 14.003 u
Reaction:
`` {}^{223}\,\mathrm{\,Ra\,}{\to }^{209}\,\mathrm{\,Pb\,}+{}^{14}\,\mathrm{\,C\,}``
`` ``
`` \,\mathrm{\,Energy\,},E=\left[m\left({}^{223}Ra\right)-\left(m\left({}^{209}Pb\right)+m\left({}^{14}\,\mathrm{\,C\,}\right)\right)\right]{c}^{2}``
`` =\left[223.018\,\mathrm{\,u\,}-\left(208.981+14.003\right)\,\mathrm{\,u\,}\right]{c}^{2}``
`` =0.034\times 931\,\mathrm{\,MeV\,}``
`` =31.65\,\mathrm{\,MeV\,}``
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Qstn #8
Show that the minimum energy needed to separate a proton from a nucleus with Z protons and N neutrons is
∆E=(MZ-1,N+MB-MZ,N)c2where M_Z_,N = mass of an atom with Z protons and N neutrons in the nucleus and M_B = mass of a hydrogen atom. This energy is known as proton-separation energy.
Ans : Given:
Mass of an atom with Z protons and N neutrons = M_Z,_N
Mass of hydrogen atom = M_H
As hydrogen contains only protons, the reaction will be given by
`` {E}_{Z\mathit{,}N}\to {E}_{\,\mathrm{\,Z\,}-1,N}+{p}_{1}``
`` \Rightarrow {E}_{Z\mathit{,}N}\to {E}_{z-1,N}{+}^{1}{\,\mathrm{\,H\,}}_{1}``
∴ Minimum energy needed to separate a proton, `` ∆E=({M}_{\,\mathrm{\,Z\,}-1,\,\mathrm{\,N\,}}+{M}_{\,\mathrm{\,H\,}}-{M}_{\,\mathrm{\,Z\,},\,\mathrm{\,N\,}}){c}^{2}``
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Qstn #9
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.
Ans : Before the separation of neutron, let the mass of the nucleus be M_Z_,N.
Then, after the separation of neutron, the mass of the nucleus will be M_Z,_N-1.
The reaction is given by
`` {E}_{Z\mathit{,}N}={E}_{Z,N-1}+{}_{0}{}^{1}\,\mathrm{\,n\,}``
If `` {M}_{\,\mathrm{\,N\,}}`` is the mass of the neutron, then the energy needed to separate the neutron b`` \left(∆E\right)`` will be
`` ∆E`` = (Final mass of nucleus + Mass of neutron - Initial mass of the nucleus)c²
`` ∆E=({M}_{Z,\,\mathrm{\,N\,}-1}+{M}_{\,\mathrm{\,N\,}}-{M}_{Z,\,\mathrm{\,N\,}}){c}^{2}``
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Qstn #10
³²P beta-decays to ³²S. Find the sum of the energy of the antineutrino and the kinetic energy of the β-particle. Neglect the recoil of the daughter nucleus. Atomic mass of ³²P = 31.974 u and that of ³²S = 31.972 u.
Ans : Given:
Atomic mass of ³²P, m(³²P) = 31.974 u
Atomic mass of ³²S, m(³²S) = 31.972 u
Reaction:
`` {\,\mathrm{\,P\,}}^{32}\to {\,\mathrm{\,S\,}}^{32}{+}_{1}{\,\mathrm{\,v\,}}^{0}{+}_{-1}{\,\mathrm{\,\beta \,}}^{0}``
Energy of antineutrino and β-particle, E = [m(³²P)`` -``m(³²S)]c²
= (31.974 u- 31.972 u)c²
= 0.002 × 931 = 1.862 MeV
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