NCERT Solutions for 12th Physics, Chapter 13 - Nuclei are available here. In this article, you will find solutions from question number 13.12 to 13.22. This article is a continuation of NCERT Solutions for CBSE Class 12 Physics ‒ Chapter 13: Nuclei (Part I) where solutions from question number 13.1 to 13.11 are available. These solutions are frequently asked in CBSE board exam 2018 and competitive exams (like JEE Main, WBJEE, NEET etc).
NCERT Solutions for CBSE Class 12 Physics ‒ Chapter 13: Nuclei from question number 13.12 to 13.22 are given below
Question 13.12: Find the Q-value and the kinetic energy of the emitted α-particle in the α- decay of (a) _{88}Ra^{226} and (b) _{86}Rn^{220}.
Given m (_{88}Ra^{226}) = 226.02540 u, m (_{86}Rn^{222}) = 222.01750 u, m (_{86}Rn^{222}) = 220.01137 u, m (_{84}Po^{216}) = 216.00189 u.
Solution 13.12
(a) Nuclear reaction:
_{88}Ra^{226} → _{86}Rn^{222} + _{2}He^{4}
Q = (M_{Ra }– M_{Rn} – M_{He}).c^{2}
= (5.297 × 10^{-3}) × (931.5) = 4.934 MeV
Kinetic energy of emitted alpha particle:
The energy liberated in the nuclear reaction displays in the form of kinetic energy of the particles of products.
Kinetic energy of products
(MV^{2}/2) + (mv^{2}/2) = 4.93 MeV …(1)
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Where,
M = mass of radon nuclei
m = mass of helium nuclei
V = velocity of radon nuclei
v = velocity of helium nuclei
To keep the momentum conserve
MV = mv
Or
M / m = v / V…………………………….(2)
This implies nothing but the simple fact that the most of the kinetic energy
(or Q calculated above) will be retained by the helium nuclei, eq. (2) pointing that it will travel with much large velocity as compared to the velocity of the radon nuclei.
Dividing eq. (1) throughout by V^{2} and making substitutions from eq. (2)
(M^{2}/m) + M = 9.86 / V^{2}
Or
V^{2} = 7.86 × 10^{-4}
KE of Rn = MV^{2} / 2 = 0.087 MeV
Therefore, KE of alpha particle = 4.93 – 0.087 = 4.85 MeV.
(b) Nuclear reaction:
_{86}Rn^{220} → _{84}Po^{216} + _{2}He^{4}
Calculating Q in a similar fashion, Q = 6.41 MeV
Kinetic energy of emitted alpha particle:
As solved in part (a) of this question
KE of products
[(MV)V / 2] + [(mv)v / 2] = 6.41
where
M = mass of polonium nuclei
m = mass of helium nuclei
V = velocity of polonium nuclei
v = velocity of helium nuclei
By the same arguments and solving technique used in part (a) of this question, V comes to be
V = 0.033
KE of Po = MV^{2} / 2 = 0.117 MeV
Therefore KE of alpha particle = 6.41 – 0.117 = 6.29 MeV
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Question 13.13: The radionuclide ^{11}C decays according to
^{11}C_{6} → _{5}B^{11} + e^{+} + v: half life = 20.3 minute.
The maximum energy of the emitted positron is 0.960 MeV.
Given the mass values:
m (_{6}C^{11}) = 11.011434 u and m (_{6}B^{11}) = 11.009305 u.
Calculate Q and compare it with the maximum energy of the positron emitted.
Solution 13.13:
Nuclear Reaction:
_{6}C^{11} → _{5}B^{11} + e^{+} + v
Δ m = [m (_{6}C^{11}) ‒ m (_{5}B^{11}) ‒ 2 m_{e}] = 0.001033 u
Now, 1 u = 931 MeV
Q = 0.001033 × 931 MeV = 0.961 MeV, which is maximum energy emitted by positron.
This Q has energy distributed as
Q = E_{d }+ E_{e} + E_{v}
For constant momentum, the energy carried by daughter nuclei and neutrino is nearly zero. The positron carries maximum energy.
Hence max E_{e }≈ Q.
Question 13.14: The nucleus _{10}Ne^{23} decays of β^{‒} emission. Write down the β-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:
m (_{10}Ne^{23}) = 22.994466 amu, m (_{11}Na^{23}) = 22.989770 amu.
Solution 13.14:
Nuclear Reaction:
_{10}Ne^{23 }→ _{11}Na^{23} + e- + n + Q
Q = [mass of Ne – mass of Na – mass of electron] × c^{2}
Using atomic masses
Q = [mass of Ne – mass of Ma].c^{2} = 4.37 MeV
For the reasons argued in previous questions, the maximum kinetic energy of electron
Maximum E_{e} = Q = 4.37 MeV.
Question 13.15: The Q value of a nuclear reaction A + b → C + d is defined by Q = [m_{A} + m_{b} – m_{C }– m_{d}]c^{2}
where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic
(i) _{1}H^{1} + _{1}H^{3} → _{1}H^{2} + _{1}H^{2}
(ii) _{6}C^{12} + _{6}C^{12} → _{10}Ne^{20} + _{2}He^{4}
Atomic masses are given to be
m(_{1}H^{2}) = 2.014102 u; m(_{1}H^{3}) = 3.016049 u; m(_{6}C^{12}) = 12.000000 u; m(_{10}Ne^{20}) = 19.992439 u
Solution 13.15:
A + b → C + d
Q = [m_{A} + m_{b} – m_{C }– m_{d}]c^{2}
(i) Nuclear Reaction:
_{1}H^{1 }+ _{1}H^{3} → _{1}H^{2} + _{1}H^{2}
Q = [-4.33 X 10^{-3}] × [931.5] = -4.03 MeV
Reaction is endothermic. This much energy has to be supplied.
(ii) Nuclear Reaction:
_{6}C^{12} + _{6}C^{12} → _{10}Ne^{20} + _{2}He^{4}
Q = [4.958 x 10^{-3}] × [931.5] = 4.62 MeV
Question 13.16: Suppose, we think of fission of a _{26}Fe^{56} nucleus into two equal fragments _{13}Al^{28}. Is the fission energetically possible? Argue by working out Q of the process.
Solution 13.16:
Given, m_{ }(_{26}Fe^{56}) = 55.93494 u, m (_{13}Al^{28}) = 27.98191 u.
Nuclear Reaction:
_{26}Fe^{56} → _{13}Al^{28} + _{13}Al^{28} + Q
Q = [mass of Fe – mass of aluminium – mass of aluminium].c.c = [-0.0288].[931.5] = ‒26.9 MeV
The reaction being endothermic is not possible.
Question 13.17: The fission properties of _{84}Pu^{239} are very similar to those of _{92}U^{235}. The average energy released per fission is 180 MeV. How much energy in MeV is released if all the atoms in 1 kg of pure _{94}Pu^{239} undergo fission.
Solution 13.17:
Given:
Energy released per fission = 180 MeV
1 g of Pu contains = Avogadro’s No. / 239
1 kg of Pu contains, N = 2.5197 x 10^{24} atoms
1 atom releases = 180 MeV
Therefore,
N atoms release = 4.536 x 10^{26} MeV.
Question 13.18: A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much _{92}U^{235} did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of _{92}U^{235} and that this nuclide is consumed only by the fission process.
Solution 13.18:
Energy generated per gram of U = (6 x 10^{23} x 200 x 1.6 x 10^{-13})/235
The amount of U consumed in 5 years with 80% on-time
= (5 x 0.8 x 3.154 x 10^{16} x 235) / (1.2 x 1.6 x 10^{13}) = 1544 kg
Thus the initial amount = 3088 kg
Question 13.19: How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as
_{1}H^{2} + _{1}H^{2} → _{2}He^{3} + n + 3.27 MeV
Solution 13.19:
Nuclear Reaction:
_{1}H^{2} + _{1}H^{2} → _{2}He^{3} + n + 3.27 MeV
No. of atoms in 2 kg deuterium
S = (Avogadro’s No. × 2000)/2 = 6.023 x 10^{26} atoms
Energy of 1 atom = 3.27 / 2 = 1.635 MeV
Therefore
Energy of S atoms = 9.85 x 10^{26 }MeV = 1.576 x 10^{14} J.
We know
Power = Energy / Time
Hence,
Time = 1.576 x 10^{12} s = About 4.9 x 10^{4} years.
Question 13.20: Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.)
Solution 13.20:
Given:
R = 2 fm
For head on collision
d = R + R = 4 fm
Height of the potential barrier, V = e^{2}/(4πϵ_{o}d)
Substituting values yields
V = 5.76 x 10^{-14} J = 360 keV.
Question 13.21: From the relation R = R_{}A^{1/3}, where R_{} is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of A).
For the β^{+ }(positron) emission from a nucleus, there is another competing process known as electron capture (electron from an inner orbit, say, the K–shell, is captured by the nucleus and a neutrino is emitted).
e^{‒} + zX^{A} → _{z‒1}Y^{A} + v
Show that if β^{+} emission is energetically allowed, electron capture is necessarily allowed but not vice–versa.
Solution 13.21:
We know
Nuclear mass density = mass / volume …(1)
Now Mass ∝ A
And Volume ∝ R^{3}
Since R ∝ A^{1/3}
Therefore
Volume ∝ A
When putting these proportional values in eq. (1), A gets cancelled. Hence proved!
Question 13.22: For the β^{+} (positron) emission from a nucleus, there is another competing process known as electron capture (electron from an inner orbit, say, the K–shell, is captured by the nucleus and a neutrino is emitted).
e^{+} + _{z}X^{A} →_{z‒1}Y^{A} + v
Show that if β^{+ }emission is energetically allowed, electron capture is necessarily allowed but not vice–versa.
The two processes:
For positron capture,
_{Z}X^{A} → _{z‒1}Y^{A} + e^{+} + v + Q_{1}
For electron capture,
e^{‒} + _{Z}X^{A} → _{z‒1}Y^{A} + v + Q_{2}
Calculating Q_{1}:
Q_{1} = [m_{N}(_{Z}X^{A}) – m_{N} (_{z‒1}Y^{A}) ‒ 2m_{e}] c^{2}
Q_{2} = [m_{N}(_{Z}X^{A}) + m_{e} ‒ m_{N }(_{z‒1}Y^{A})]c^{2}
From the above calculations, it is obvious that, if Q_{1} > 0 then Q_{2} > 0
But if Q_{2 }> 0, then it can’t be said that Q_{1} > 0
Hence proved!
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