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 Bihar Board Class 12 Physics Solutions Chapter 13 Nuclei

Bihar Board Class 12 Physics Nuclei Textbook Questions and Answers

You may find the following data useful in solving the exercises:
e = 1.6x 10-19C,
1(4πε0) = 9 x 109Nm2/C2,
I MeV = 1.6 x 10-13J,
1 year = 3.154 x 107s
mH = 1.007825 u,
m (42He) = 4.002603 u,

N = 6.023 x 1023 per mole
k = 1.381 x 10-23 J0K-1
lu = 931.5 MeV/c2
mn = 1.008665 u
me = 0.000548 u

Question 1.
(a) Two stable isotopes of lithium 63Li and 73Li have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512u and 7,01600u, respectively. Find the atomic weight of lithium.
(b) Boron has two stable isotopes, 105B and 115B Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of 105B and 115B.
Answer:

Let A be the atomic weight of Lithium.
∴ A is given by the formula,

(b) Here. A = Atomic weight of B = 10.811 u.
Let % and (100 – a)%betheabundancesof 105B and 115B isotopes respectively

Question 2.
The three stable isotopes of neon : 2010Ne 2110Ne and 2210Ne have respective abundances of 90.51 %, 0.27 % and 9.22 %. The atomic masses of three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon.
Answer:

Question 3.
Obtain the binding energy in MeV of a nitrogen nucleus (147N), given m (147N) = 14.00307 u.
Answer:
147N is made up of 7 protons and 7 neutrons.
Here,
mH = 1.00783 u
mn = 1.00867 u
mN = 14.00307 u
B.E. = ?
∴ Mass of nucleons forming nucleus.
= 7mH + 7mn
= 7 x 1.00783 u + 7 x 1.0086 7 u
= 7.05481 + 7.06069 = 14.11550 u.
If ∆m be the mass defect, then
∆m = mass of nucleons – mass of nucleus
= 14.11550 u -14.00307 u
= 0.11243 u

Using Formula,
B.E. = ∆m x 931 MeV
= 0.11243 x 931 MeV = 104.67 MeV = 104.7 MeV.

Question 4.
Obtain the binding energy of the nuclei 5626Fe and 20983Bi in units of MeV from the following :
m (5626Fe) = 55.934939 u m (20983Bi) = 208.980388 u
Answer:
Here, mass of hydrogen atom, mH 1.007825 a.m.u.
mass of neutron, mn = 1.008665 a.m.u.

(i) 5626Fe nucleus contain 26 protons and (56 – 26) = 30 neutrons
∴ Mass of 26 protons = 26 x 1.007825 am.u = 26.20345 a.m.u.
Mass of 30 neutrons = 30 x 1.008665 am.u = 30-25995 a.m.u.
Total mass of 56 nucleons = (26.20345 + 30.25995) a.m.u
= 56.46340 a.m.u.
Mass of 5626Fe nucleus = 55.934939 a.m.u.
∆m = mass defect is given by
∆m = mass of nucleons – mass of nucleus of 5626Fe.
= (56.46340 – 55.934939) a.m.u
= 0.528461 a.m.u.
Total binding energy = 0.528461 a.m.u
= 0.528461 x 931.5 MeV = 492.26 MeV.

(ii) 20983Bi nucleus contains 83 protons and (209 – 83) = 126 neutrons.
∴ Total mass of nucleons = 83 x mH +126 mn
= 83 x 1.007825 + 126 x 1.008665 = 83.649475 +127.091790 a.m.u. = 210.741260 a.m.u.
Mass of 20983B nucleus = 208.980388 a.m.u.
∴ Mass defect is given by
∆m = mass of nucleons – mass of nucleus of 20983Bi
= (210.741260 – 208.980388) a.m.u. or ∆m = 1.760872 a.m.u.
∴ B.E. of 20983Bi = 1.760872 x 931.5 MeV = 164026 MeV
Average B.E/nucleon = 1640.26209 – MeV
= 7.84MeV/nucleon

Question 5.
A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of 6329Cu atoms (of mass 62.92960 u).
Answer:
Mass of Copper coin = 3g
Symbol of Copper nucleus ZXA = 29Cu63
A copper nucleus has 29 protons each of mass 1.00783 u and it has 63 – 29 = 34 neutrons each of mass 1.00867 u
In 63 g of copper, No. of copper atoms = Avogadro Number
= N = 6.023 x 1023
∴ No. of atoms in 1 g = 6.023×102363
and No. of atoms in 1 g = 6.023×102363×3
For 1 atom of copper, mass of nucleons = Mass of 29 protons + Mass of 34 neutrons
= (29 x 1.00783 + 34 x 1.00862) amu
Mass of Copper nucleus = 62.9260 u
Mass defect, ∆m = 29 x 1.00783 + 34 x 1.00862 – 62.9260 amu (for 1 copper nucleus)
In 3 gram of copper, no. of copper nuclei = 2.868 x 1022
Mass defect for 3 gram of copper
= (29 x 1.00783 + 34 x 1.00862 – 62.9260) x 2.868 x 1022
= (83.82 + 98.36 -180.48) x 1022 amu
But 1 amu = 9.311 MeV
∴ Energy equivalent to mass defect = 1.70 x 1022 x 931
= 1582.7 x 1022 MeV
This much energy is required to break copper nuclei into constituent nucleons (protons and neutrons)
∴ Energy Required = 1582.7 x 1022 MeV.
= 1.583 x 1025 MeV = 1.583 x 1025 x 1.6 x 10-13J = 2.533 x 1012 J.

Question 6.
Write nuclear reaction equations for
(i) α – decay of 22688Ra
(ii) α – decayof 24294Pu
(iii) β – decay of 3215P
(iv) β-decay of 21083Bi
(v) β+ – decay of 116 C
(vi) β+ – decay of 9743 Tc
(vii) Electron capture of 12054Xe
Answer:
α – decay of Radium.
(i) α – decay of 22688Ra → 22286Rn + 42He (α-particle).
Parent nucelus decays into radn.

(ii) α – decay of Pu.
24294Pu → 23892U + 42He (α-particle).

(iii) β– – decay of P
3215P → 3216S + -1e° + v–
i.e.,β-decay is accompanied by relese of antineutrino.

(iv) β– – decay of 21083Bi
21083Bi → 21084P + –1e° + v–

(v) β+ – decay of 116C :
116C → 5C11 + +1e° + v
P+-decay of is accompanied by the release of neutrino.

(vi) β+ – decay of 9743Tc :
9743Tc → 9742Mo + +1e° + v

(vii) Nuclear reaction equation for electron capture of 12054Xe is given by
12054Xe + +1 e0 → 12053I + v

Question 7.
A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to a) 3.125 %, b) 1 % of its original value?
Answer:
Let N0 = initial activity
N = activity after time
T = half-life of the radioactive isotope
λ = disintegration constant
∴ initial activity A0 is given by
A0 = λ N0
Also we know that
N = N0 ((12)tT)

(a) Here A = 3.125 % of A0 = 3.125100 A0
= 3125100×1000 A0
= 132 A0 ………….(1)

∴ Activity, A = – dNdt = λ N = λN0((12)tT)
∴ A = A0 ((12)tT) ……….(2)
∴ from (1) and (2) we get,
132 = ((12)tT)
or 1325 = ((12)tT)
or 5 = tT
or t = 5 T years.
After time 5 half periods activity reduces to 3.125 % of initial activity.

(b) Here, A = 1 % of A0 = 1100A0
or AA0=1100 ………..(1)
Also we know that AA0 = e-λt …………(2)

∴ From (1) and (2), we get
1100 = e-λt or eλt = 102. 102
Taking log on both sides, we get
λt = log e102 = 2 x 2.303 log1010 = 4.606
∴ t = 4.606λ ………..(3)
Also we know that, λ = 0.693 T …….(4)

∴ From (3) and (4),,we get
t = 4.606(0.693 T)=4.6060.693 T
= 6.65 T years.

Question 8.
Thenormalactivity of living carbon – containing matter is found to be abouil5 decays percmmrte for every gram of carbon. This activity arises from the small proportion of radioactive 146C present with the stable carbon isotope 146C. When the organism is dead, its interaction with the atmosphere (which maintains -the above equilibrium activity) ceases and its activity begins to drop. From the known half-life (5,730 years) of 146C, and the measured activity, the age of the specimen can be estimated. This is the principle of 146C dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus Valley Civilisation.
Answer:
`Here, let the no. of C14 atoms per gram were N0 at t = 0 (initially), when its activity was 15 decays perminute per gram. Today after time t the no. of C14 atom per gram left is N and it shows an activity of a decays per minute per gram.
i.e., normal activity, R0 = 15 decays / min.
Present activity, R – 9decays/min.
T12 = 5,730 years
Aget = ?
Also we know that activity is directly proportional to the number of radioactive atoms, so NN0=RR0=915
Also we knoe that
NN0 = e-λt
or 915= e-λt
or e-λt = 53 = 1.6667
λt logee = loge 1.6667
λt = 2.303 log10 1.6667
t = 0.5109λ …………(1)
Also we know that decay constant is given by
λ = 0.693 T12=0.6935,730 ………..(2)
∴ From (1) and (2), we get
t = 0.51090.693 x 5,730 = 4225.15 years.

Question 9.
Obtain the amount of 6027Co necessary to provide a radioactive source of 8.0 mCi strength. The half-life of 146Co is 5.3 years.
Answer:
Here, R = activity = strength of radioactive source = 8.0 m Ci = 8 x 10-3 x 3.7 x. 1010 disintegrations/sec.
= 2.96 x 108 disintegrations/sec (∵ 1 Ci = 3.7 x 1010 disintegrations s-1).
T12 = half-life of 6027Co = 5.3 years
s = 5.3 x 365 x 24 x 60 x 60 s = 1.67 x 108 s.
Nt = no of nucli present at time = ?
Using relation,
R = – dNdt = – ddt(N0e-λt)
= + λ N0e-λt

Also know that
λ = 0.693 T1=0.6931.67×108 s
∴ From (1) and (2), we get
0.6931.67×108 Nt = 2.96 x 108 1.67 x10s 1
Nt = 2.96×108×1.67×1080.693
= 7.133 x 1016
A mount of 146Co = ?
We know that 60 g of Cobalt contains 6.023 x 1023 cobalt atoms or nuclei.
i e., 6.023 x 1023 atoms of Co have mass = 60 g
1 atom of Co have mass = 606.023×1023 g
∴ 7.133 x 1016 atoms of Co have mass = 606.023×1023 x 7.133 x 1016
i.e.,Amount of 6027C0 needed = 7.11 μg.

Question 10.
The half-life of 9038Sr is 28 years. Whal is the ‘ disintegration rate of 15 mg of this isotope?
Answer:
Here, T1/2 = Half-life of 9038Sr = 28.0 years = 28 x 365 x 24 x 3,600 s = 88.3 x 107 s.
Amount of 9038Sr = 15 mg = 15 x 10-3g.
Now 90 gm of Sr has atoms = 6.023 x 1023.
∴ 1 g of Sr will have atom = 6.023 x 1023
∴ Number of atoms of 30Sr in 15 mg is given by
= 6.023×102390 x 15 x 10-3
= 1.004 x 1020
∴ Nt = 1.004 x 1020
Also using λ = 0.693 T12 we get
λ = 0.69388.3×107 s-1
R = ?
Using the relation, R = XN1, we get
R = 0.69388.3×107 x 1.004 x 1020
= 7.879 x 10-3 x 1013
= 7.879 x 1010 disinteg rafions s-1
= 7.879 x 1010Bq.
= 7.879×10103.7×1010 Ci (∵ 1 Ci = 3.7 x 1010 Bq)
= 2.13 Ci
∴ Disintegration rate = 2.13 Ci or 7.879 x 1010Bq.

Question 11.
Obtain approximately the ratio of the nuclear radii of the gold isotope 19779Au and the silver isotope 19747Ag .
Answer:
We know that for a spherical nucleus, its radius is given by
R = R0 AA13 ………(1)
Where A is the mass number of the nucleus, RQ is an empirical constant.
Let R1, A1 be the nuclear radius and mass number of the 19779Au isotope.
Also let R2, A2 be the nuclear radius and mass number of 19747Ag isotope.
∴ A1 = 197, A2 = 107
∴ Using (1)
R1 = A0(197)13 ………(2)
And R2 = A0(197)13 ……….(3)
Dividing (2) by (3), we get

Question 12.
Find the Q-value and the kinetic energy of the ifted α-particle in the α-decay of (a) 22688Ra (b) 22086Rn
Given m (22688Ra) 226.02540 u,m (22688Rn) = 222.01750 u
m (22288Rn) = 220.01137 u, m (21684po) 216.00189 u.
Answer:
(a) The process of a-decay of 22688Ra can be expressed as :
22688Ra -» 22286RaRn + 24He +Q ……….(1)
Q-value of the reaction is given by

When mN represents the nuclear masses of the respective nuclei.
In forms of atomic masses, eqn. (2) can be expressed as :
i
The K.E. of the emitted α-particle is given by
(K-E-)α = A−4A Q
Here, A = 226, ∴ A – 4 = 226 – 4 = 222, = 4.9342 MeV
∴ (K.E.) α = 222226 x 4.9342 MeV
= 4.8469 MeV
= 4.85 MeV.

(b) The process of α-decay of 22086Rn can be expressed as :
22086Rn → 21684Po +αHe + Q ………..(1)
The Q-value of the reaction is given by :

Let Eα be the K.E. of the a-particle emitted.
∴Eα is given by
Eα = A−4 A Q
Where A = 220, A – 4 = 216, Q = 6.41 MeV.
∴Eα is given by :
E = 216220 = 6.41 MeV.
= 6.289
= 6.29 MeV.

Question 13.
The radionuclide 11C decays according to
116C → 115B + e+ + v : T1/2 = 20.3 min
The maximum energy of the emitted positron is 0.960 MeV.
Given the mass values :
m (116C) = 11.011434 u and m (116B) = 11.009305 u,
calculate Q and compare it with the maximum energy of the positron emitted.
Answer:
The equation for the decay process is –
116C → 115B + +1e0 + v + Q
Where Q – value of the reaction or K.E. of α-particle. = mass defect in a.m.u.
= (Mass of 116C nucleus – Mass of 115B nucleus – mass of positron in a.m.u
= ⌊mN(116C)−mN(115 B)−me⌋ ………..(1)
Where mN stands for the nuclear mass of the element or particle.
∴ As 116C atom is made up of 116C nucleus and 6 electrons.
∴ Mass of 116C nucleus = mass of 116C atom – mass of 6 electrons
= 11 011434 a.m.u – 6 me ………..(2)
Similarly mass of 115B nucleus = mass of 115B atom – mass of 5 electrons.
– 11.009305 a.m.u – 5 me …….(3)

∴ From (1), (2) and (3), we get
Q = (11.011434 a.m.u. – 6me) – (11.009305 a.m.u. – 5me) -me]
= (11.011434 -11.009305) a.m.u. – 2me
= (11.011434 -11.009305) a.m.u. – 2 x 0.0005048 a.m.u.
= 0.001033 a.m.u.
= 0.001033 x 931.5 MeV = 0.962 MeV ……….(4)
Maximum kinetic energy of emitted position (β+) is 0.960 MeV ……..(5)

∴ From (4) and (5), we conclude that the Q-value of the reaction is comparable to the actual energy released in the decay process.
Q = ED + Ee + EV
Tire daughter nucleus is too heavy compared to e+ and v, so it carries neglible energy (Ed ≈ 0). If the kinetic energy (Ev) carried by the neutrino is minimum (i.e., zero), the psotron carries maximum energy, and this is practically all energy Q; hence maximum Ee ≈ Q).

Question 14.
The nucleus 2310Ne decays by β– emission. Write down the β-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that :
m (2310Ne) = 22 994466 u
m (2310Na) = 22.089770 u.
Answer:
Here,mass of 2310Ne atom = 22,094466 u.
mass of 2311Na atom = 22 989770 u.
The β-decay equation of 2310 Ne is
2310 Ne → 2311 Ne + -1e° + v + EB
Where EB is the maximum kinetic energy of the emitted p-particle and is equal to the Q-value of the reaction and is given by

where mN (2310Ne) and mN (2311Na) are the nuclear masses of 2310Ne and 2311Na respectively. If m (2310Ne) and m (2311Na)be the atomic masses of 2310Ne and 2311Na respectively, then
mN (2310Ne) = m (2311Ne) – 10 me (∴ atom of Ne contains 10 electrons)
and mN (2311Na) = m (2311Na) – 11 me ( ∴atom of Na contains 11 electrons)
∴ From (1), Eβ = [ m(2311Ne) – 1oMe] – (2311Na – 11me] – me] in a.m.u

The energy released is shared by 2311Na nucleus and the electron -neutrino pair released. As electron-neutron pair is much lighter than 2311Na nucleus, practically whole of the energy released is carried by electron-neutrino pair. When neutrino gets zero energy, the electron will carry the maximum energy. So the maximum K.E. of the electron emitted is 4.374 MeV.

Question 15.
The Q value of a nuclear reaction A + b → C + d is defined by Q = [mA + mb + mC + md]C2 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) 11H + 31H → 21H + 21H
(ii) 126C + 126C → 2010Ne + 42He
Atomic masses are given to be
m (21H) = 2.014102 u
m (31H) = 3.016049 u
m (126C) = 12.000000 u
m (2010 Ne) = 19.992439 u
Answer:
(i) The given reaction is :
11H + 31H → 21H + 21H
∴ Q-value of the reaction is :

Where mN refers to nuclear masses.
m (11H) = mN (11H) + me or mN (11H) = m(11H) – me.
Similarly, mN (31H) = m(31H) – me

= 1.007825 + 3.016049 – 2 x 2.014102
= – 0.004339 a. m.u.
= -0.00433 x 1.66 x 10-27 kg.
∴ From (1), Q = – 0.00433 x 1.66 x 10-27 x (3 x 108)2
= – 6.46 x 10-13 J
= −6.46×10−131.6×10−13 MeV = – 4.03 eV
Since the Q-value is negative, so the reaction is endothermic,

(ii) The given reaction is :

Q-value of the reaction is given by

Where mN refers to masses of respective nuclei and

∴ From (2)

Since the Q-value is positive, so the reaction is exothermic.

Question 16.
Suppose, we think of fission of a 5626Fe nucleus into two equal fragments, 2813 Al . Is the fission energetically possible? Argue by working out Q of the process. Given m (5626Fe) = 55.93494 u and m (2813 Al) = 27.98191 u.
Answer:
The given reaction for the decay process is :

where Q-value of the reaction is given by.
Q = [mn + m(5626Fe) – 2m(2813Al)] x 931.5 MeV
= [1.00867 + 55.93494 – 2 x 27.98191] x 931.5 MeV
= (56.94361 – 55.96382) x 931.5
= 0.97979 x 931.5 MeV
= 912.67 MeV.
Although the Q-value is positive but the mass number (i.e., A) is not conserved in this process. Therefore the process of fission of 5626Fe cannot be possible.

Question 17.
The fission properties of 23994Pu are very similar to those of 23592U. 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 23994Pu undergo fission?
Answer:
Here, average energy released per fission of 23994pu = 180 MeV.
Total energy released (MeV) = ?
Amount of 23994pu = 1 kg = 1,000 g
We know that 239 g of 23994pu contains = 6.023 x 1023 fissionable nuclei.
or 1g of 23994pu contains = 6.023×1023239
∴ No. of fissionable nuclei in 1,000 g of 23994pu
6.023×1023239 x 1,000
= 25.2 x 1023 nuclei.
∴ Total energy released in fission of 1 kg of Pu
= 180 x 25.2 x 1023 MeV
= 4536 x 1023 MeV
= 4.536 x 1026 MeV.

Question 18.
A1000 MW fission reactor consumes half of its fuel in 5.0 y. How much 235 JJ did it contain initially? Assume that the reactor operates 80 % of the time, that all the energy generated arises from the fission of 23592U and that this nuclide is consumed only by the fission process.
Answer:
Here, P = power of reactor = 1,000 MW = 103 x 106 W = 109 Js-1
t = time = 5 years = 5 x 365 x 24 x 3,600 s = 1.577 x 108 s.
If E be the amount of energy delievered by the reactor when it operates with 80 % of time, then

or m’ = half of the fuel consumed in 5 g.
∴ If M be the initial mass of 23592U, then
m = 2m’
= 2×1544.15
= 3088.3 kg.

Question 19.
How long can an electric lamp of 100 W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as ?
21H + 21H → 32He + n + 3.27 MeV
Answer:
Here, P – Power of lamp = 100 W.
m = mass of deuterium = 2.0 kg.
The fusion reaction is given by
21H + 21H → 32He + n + 3.27 MeV ……..(1)
From eq. (1), it is clear that 2 nuclei of 21H (deuterium) combine to give 3.27 MeV of energy.
Now 2 g of 21H contains 6.023 x 1023 nuclei.
∴ Numer of 21H nuclei in 2 kg of deterium = 6.023×10232×10−3 x 2
= 6.023 x 1026 nuclei.
Energy released in fusion of two nuclei of 21H = 3.27 MeV.
∴ Energy released in fusion of 6.023 x 1026 nuclei of it is given by
E = 3.27×6.0232×1026 MeV
or E = 3.27×6.0232×1026 x 1026 x 1.6 x 10-13 J = 15.76 x 1013J ………(2)
Let t = time for which the electric lamp can glow due to this energy = ?
∴ energy consumed by the lamp, E’ = Pt = 100 t ……….(3)
∴ According to law of conservation of energy
E’ = E

Question 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.)
Answer:
Let r be the distance of closest approach of the two deutrons for head-on-collision.
Here r = 2 fm = 2 x 10-15 m.
e = charge on each deutron
= 1.6 x 10-19 C.

Let E be the initial mechanical energy of the two deutrons before collision.
∴ E = 2 K.E. …………(1)
where K.E. = Kinetic energy of each deutron. When the two deutrons stop, their energy is totally electric potential energy (= U) given by
U = Fe. r

∴ according to the law of conservation of energy,
E = U

∴ height of potential barrier = K.E. = 360 KeV.

Question 21.
From the relation R = R0 A1/3, where R0 is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (t.e., independent of A).
Answer:
The expression of the radius of nucleus is given by
R = R0 A1/3 ………..(1)
Where R0 is a constant and A is the mass number of a nucleus. This can be used to calculate the density of the nucleus, mass of the nucleus of the atom = A a.m.u.
= A x 1.66 x 10-27 kg.
If V be the volume of the nucleus, then

Thus from equation (4) we see that ρ is independent of A, hence we conclude that p is nearly constant for all nuclei.

Question 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).

Show that if β+ emission is energetically allowed, electron capture is necessarily allowed but not vice-versa.
Answer:
The β+ emission from a nucleus AZXcan be represented as :

The other competing process of electron capture may be represented as :

The energy released Qj in Eq. (1) is given by


i.e., If positron emission is energetically allowed, electron capture is necessarily allowed. But Q2 > 0 does not necessarily mean Q1> 0. Hence the reverse is not true.