NCERT Solutions for Class 12 Physics Chapter 13 Nuclei

NCERT Solutions for Class 12 Physics Chapter 13 – Free PDF Download

The NCERT Solutions for Class 12 Physics Chapter 13 Nuclei is provided here to help students in their CBSE Class 12 second term exam preparations. All the NCERT Solutions for Class 12 Physics are prepared by our subject experts as per the latest term – II CBSE Syllabus for 2021-22 and its guidelines. Students can make the best of their preparations by referring to these solutions.

The NCERT Solutions for Class 12 Physics of this chapter provided here has answers to the textbook questions along with Class 12 important questions, exemplary problems, worksheets and exercises. All these will help students gain complete knowledge on the topics Nuclei. Students can avail free PDF of the NCERT Solutions for Class 12 Physics Chapter 13 by downloading it from the link given below.

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Class 12 Physics NCERT Solutions for Chapter 13 Nuclei Important Questions


Q 13.1 (a) Lithium has two stable isotopes 36Li_{3}^{6}{Li} and 37Li_{3}^{7}{Li} have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium.

(b) Boron has two stable isotopes, 510B_{5}^{10}{B} and 511B_{5}^{11}{B} . 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 510B_{5}^{10}{B} and 511B_{5}^{11}{B}.

A1 :

 (a) Mass of 36Li_{3}^{6}{Li} lithium isotope , m1 = 6.01512 u

Mass of 37Li_{3}^{7}{Li} lithium isotope , m2 = 7.01600 u

Abundance of 36Li_{3}^{6}{Li} , n1= 7.5%

Abundance of 37Li_{3}^{7}{Li} , n2= 92.5%

The atomic mass of lithium atom is:

m=m1n1+m2n2n1+n2m =\frac{ m_1 n_1 + m_2 n_2 }{ n_1 + n_2 }

 

m=6.01512×7.5+7.01600×92.57.5+92.5m =\frac{ 6.01512 \times 7.5 + 7.01600 \times 92.5 }{ 7.5 + 92.5 }

= 6.940934 u

(b) Mass of boron isotope 510B_{5}^{10}{B} m1 = 10.01294 u

Mass of boron isotope 511B_{5}^{11}{B} m2 = 11.00931 u

Abundance of 510B_{5}^{10}{B}, n1 = x%

Abundance of 511B_{5}^{11}{B}, n2= (100 − x)%

Atomic mass of boron, m = 10.811 u

The atomic mass of boron atom is:

m=m1n1+m2n2n1+n2m =\frac{ m_1 n_1 + m_2 n_2 }{ n_1 + n_2 }

 

10.811=10.01294×x+11.00931×(100x)x+100x10.811 =\frac{ 10.01294 \times x + 11.00931 \times ( 100 – x ) }{ x + 100 -x }

1081.11 = 10.01294x + 1100.931 – 11.00931 x

x = 19.821/0.99637 = 19.89 %

And 100 − x = 80.11%

Hence, the abundance of 510B_{5}^{10}{B} is 19.89% and that of 511B_{5}^{11}{B} is 80.11%.

 Q 13.2 :The three stable isotopes of neon: 1020Ne_{10}^{20}{Ne}, 1021Ne_{10}^{21}{Ne} and 1022Ne_{10}^{22}{Ne} have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon.

Answer

Atomic mass of 1020Ne_{10}^{20}{Ne}, m1= 19.99 u

Abundance of 1020Ne_{10}^{20}{Ne}, n1 = 90.51%

Atomic mass of 1021Ne_{10}^{21}{Ne}, m2 = 20.99 u

Abundance of 1021Ne_{10}^{21}{Ne}, n2 = 0.27%

Atomic mass of 1022Ne_{10}^{22}{Ne}, m3 = 21.99 u

Abundance of 1022Ne_{10}^{22}{Ne}, n3 = 9.22%

Below is the average atomic mass of neon:

m=m1n1+m2n2+m3n3n1+n2+n3m=\frac{ m_1 n_1 + m_2 n_2 + m_3 n_3 }{ n_1 + n_2 + n_3 }

= 19.99×90.51+20.99×0.27+21.99×9.2290.51+0.27+9.22\frac{ 19.99 \times 90.51 + 20.99 \times 0.27 + 21.99 \times 9.22 }{ 90.51 + 0.27 + 9.22 }

= 20.1771 u

Q 13.3: Obtain the binding energy in MeV of a nitrogen nucleus 714N_{7}^{14}{N}, given m(714N_{7}^{14}{N})=14.00307 u

Ans:

Atomic mass of nitrogen 714N_{7}^{14}{N}, m = 14.00307 u

A nucleus of 714N_{7}^{14}{N}nitrogen contains 7 neutrons and 7 protons.

∆m = 7mH + 7mn − m is the mass defect the nucleus

Where,

Mass of a proton, mH = 1.007825 u

Mass of a neutron, mn = 1.008665 u

∆m = 7 × 1.007825 + 7 × 1.008665 − 14.00307

= 7.054775 + 7.06055 − 14.00307

= 0.11236 u

But 1 u = 931.5 MeV/c2

∆m = 0.11236 × 931.5 MeV/c2

Eb = ∆mc2 is the binding energy of the nucleus

Where, c =Speed of light

Eb=0.11236×931.5(MeVc2)×c2E_b = 0.11236 \times 931.5 \left ( \frac{MeV}{c^{2}} \right ) \times c^{2}

= 104.66334 Mev

Therefore, 104.66334 MeV is the binding energy of the nitrogen nucleus.

Q 13.4: Obtain the binding energy of the nuclei 2656Fe_{26}^{56}{Fe} and 83209Bi_{83}^{209}{Bi} in units of MeV from the following

data:

m (2356Fe_{23}^{56}{Fe}) = 55.934939 u

m(83209Bi_{83}^{209}{Bi}) = 208.980388 u

Ans :Atomic mass of 2656Fe_{26}^{56}{Fe}, m1 = 55.934939 u

2656Fe_{26}^{56}{Fe} nucleus has 26 protons and (56 − 26) = 30 neutrons

Hence, the mass defect of the nucleus, ∆m = 26 × mH + 30 × mn − m1

Where,

Mass of a proton, mH = 1.007825 u

Mass of a neutron, mn = 1.008665 u

∆m = 26 × 1.007825 + 30 × 1.008665 − 55.934939

= 26.20345 + 30.25995 − 55.934939

= 0.528461 u

But 1 u = 931.5 MeV/c2

∆m = 0.528461 × 931.5 MeV/c2

Eb1E_{b_{1}} = ∆mcis the binding energy of the nucleus.

Where, c = Speed of light

Eb1=0.528461×931.5(MeVc2)×c2E_{b_{1}} = 0.528461 \times 931.5 \left ( \frac{MeV}{c^{2}} \right ) \times c^{2}

= 492.26 MeV

Average binding energy per nucleon = 492.2656=8.79MeV\frac{492.26}{56} = 8.79 MeV

Atomic mass of 83209Bi_{83}^{209}{Bi}, m2 = 208.980388 u

83209Bi_{83}^{209}{Bi} nucleus has 83 protons and (209 − 83) 126 neutrons.

The mass defect of the nucleus is given as:

∆m’ = 83 × mH + 126 × mn − m2

Where,

Mass of a proton, mH = 1.007825 u

Mass of a neutron, mn = 1.008665 u

∆m’ = 83 × 1.007825 + 126 × 1.008665 − 208.980388

= 83.649475 + 127.091790 − 208.980388

= 1.760877 u

But 1 u = 931.5 MeV/c2

∆m’ = 1.760877 × 931.5 MeV/c2

Eb2 = ∆m’c2 is the binding energy of the nucleus.

= 1.760877 × 931.5 (MeVc2)×c2 \left ( \frac{MeV}{c^{2}} \right ) \times c^{2}

= 1640.26 MeV

Average binding energy per nucleon = 1640.26/209 = 7.848 MeV

Q 13.5: 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 2963Cu_{29}^{63}{Cu} with mass = 62.92960 u.

Ans :

Mass of a copper coin, m’ = 3 g

Atomic mass of 2963Cu_{29}^{63}{Cu} atom, m = 62.92960 u

The total number of 2963Cu_{29}^{63}{Cu} atoms in the coin, N=NA×mMass  numberN = \frac{N_A \times m’}{Mass \; number}

Where,

NA = Avogadro’s number = 6.023 × 1023 atoms /g

Mass number = 63 g

N=6.023×1023×363N = \frac{6.023 \times 10^{23}\times 3}{63} = 2.868 x 1022 atoms

2963Cu_{29}^{63}{Cu} nucleus has 29 protons and (63 − 29) 34 neutrons

Mass defect of this nucleus, ∆m’ = 29 × mH + 34 × mn − m

Where,

Mass of a proton, mH = 1.007825 u

Mass of a neutron, mn = 1.008665 u

∆m’ = 29 × 1.007825 + 34 × 1.008665 − 62.9296

= 0.591935 u

Mass defect of all the atoms present in the coin, ∆m = 0.591935 × 2.868 × 1022

= 1.69766958 × 1022 u

But 1 u = 931.5 MeV/c2

∆m = 1.69766958 × 1022 × 931.5 MeV/c2

Eb = ∆mc2 is the binding energy of the nuclei of the coin

= 1.69766958 × 1022 × 931.5 MeV/c× c2

= 1.581 × 1025 MeV

But 1 MeV = 1.6 × 10−13 J

Eb = 1.581 × 1025 × 1.6 × 10−13

= 2.5296 × 1012 J

Therefore, the energy required to separate all the neutrons and protons from the given coin is  2.5296 × 1012 J

Q 13.6: Write nuclear reaction equations for

(i) α-decay of88226Ra_{88}^{226}{Ra}  (ii) α-decay of94242Pu_{94}^{242}{Pu}

(iii) β-decay of 1532P_{15}^{32}{P} (iv) β -decay of83210Bi_{83}^{210}{Bi}

(v) β+-decay of611C_{6}^{11}{C} (vi) β+ -decay of4397Tc_{43}^{97}{Tc}

(vii) Electron capture of54120Xe_{54}^{120}{Xe}

Ans:

In helium, α is a nucleus 24He_{2}^{4}{He} and β is an electron (e for β and e+ for β+). 2 protons and 4 neutrons is lost in every α decay. Whereas 1 proton and a neutrino is emitted from the nucleus in every β + decay. In every β decay, there is a gain of 1 proton and an anti-neutrino is emitted from the nucleus.

For the given cases, the various nuclear reactions can be written as:

(i) 88226Ra  86222Rn  +  24He_{88}^{226}{Ra} \rightarrow\; _{86}^{222}{Rn} \; + \; _{2}^{4}{He}

(ii) 94242Pu  92238U  +  24He_{ 94 }^{ 242 }{ Pu } \rightarrow\; _{ 92 }^{ 238 }{ U } \; + \; _{ 2 }^{ 4 }{ He }

(iii) 1532P  1632S  +  e+vˉ_{15}^{ 32 }{P} \rightarrow\; _{ 16 }^{ 32 }{ S } \; + \; e^- +\bar{v}

(iv) 83210B  84210PO  +  e+vˉ_{83}^{ 210 }{B} \rightarrow\; _{ 84 }^{ 210 }{ PO } \; + \; e^- +\bar{v}

(v) 611C  511B  +  e++  v_{6}^{ 11 }{C} \rightarrow\; _{ 5 }^{ 11 }{ B } \; + \; e^+ + \; v

(vi) 4397Tc  4297MO  +  e++  v_{43}^{ 97 }{Tc} \rightarrow\; _{ 42 }^{ 97 }{ MO } \; + \; e^+ + \; v

(vii) 54120Xe+  e+  53120I  +  v_{54}^{ 120 }{Xe}+ \; e^+ \rightarrow\; _{ 53 }^{ 120 }{ I } \; + \; v

Q 13.7: 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?

Ans:

The half-life of the radioactive isotope = T years

Nis the actual amount of radioactive isotope.

(a) After decay, the amount of the radioactive isotope = N

It is given that only 3.125% of N0 remains after decay. Hence, we can write:

NN0=3.125%=3.125100=132\frac{N}{N_0} = 3.125 \% = \frac{3.125}{100}=\frac{1}{32}

But,  NN0=eλt\frac{N}{N_0} = e^{-\lambda t}

Where, λ = Decay constant

t = Time

eλt=132e^{-\lambda t} = \frac{1}{32}

 

λt=ln1ln32-\lambda t = ln 1 – ln 32

 

λt=03.4657-\lambda t = 0 – 3.4657

Since, λ=0.693T-\lambda = \frac{0.693}{T}

t = 3.4660.693T  5Tyears\frac{3.466}{\frac{0.693}{T}} \approx \; 5 T years

Hence, the isotope will take about 5T years to reduce to 3.125% of its original value.

(b) After decay, the amount of the radioactive isotope = N

It is given that only 1% of N0 remains after decay. Hence, we can write:

NN0=1%=1100\frac{N}{N_0} = 1 \% = \frac{1}{100}

 

NN0=eλt\frac{N}{N_0} = e^{-\lambda t}

 

eλt=1100e^{-\lambda t} = \frac{1}{100}

 

λt=ln1ln100-\lambda t = ln1 – ln100

 

eλt=e^{-\lambda t} = 0 – 4.6052

t = 4.6052/λ\lambda

Since, λ = 0.693/T

t=4.60520.693Tt = \frac{4.6052}{\frac{0.693}{T}} = 6.645 T years

Hence, the isotope will take about 6.645T years to reduce to 1% of its original value.

 Q 13.8: The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive614C_{6}^{14}{C}  present with the stable carbon isotope 612C_{6}^{12}{C} . 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 (5730 years) of614C_{6}^{14}{C}, and the measured activity, the age of the specimen can be approximately estimated. This is the principle of 614C_{6}^{14}{C} 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.

Ans :

Decay rate of living carbon-containing matter, R = 15 decay/min

Let N be the number of radioactive atoms present in a normal carbon-containing matter.

Half life of 614C,T12_{6}^{14}{C},T_{\frac{1}{2}} = 5730 years

The decay rate of the specimen obtained from the Mohenjo-Daro site:

R’ = 9 decays/min

Let N’ be the number of radioactive atoms present in the specimen during the Mohenjo-Daro period.

Therefore, the relation between the decay constant, λ and time, t is:

NN=RR=eλt\frac{N}{N’}=\frac{R}{R’}= e^{-\lambda t}

 

eλt=915=35e^{-\lambda t} = \frac{9}{15}=\frac{3}{5}

 

λt=loge35=0.5108-\lambda t = log_e\frac{3}{5} = -0.5108

 

t=0.5108λt = \frac{0.5108}{\lambda}

But λ=0.693T12=0.6935730{\lambda} = \frac{0.693}{T_\frac{1}{2}}=\frac{0.693}{5730}

 

t=0.51080.6935730=4223.5yearst = \frac{0.5108}{\frac{0.693}{5730}}= 4223.5 years

So, the approximate age of the Indus-Valley civilization is 4223.5 years.

Q 13.9: Obtain the amount of 2760Co_{27}^{60}{Co} necessary to provide a radioactive source
of 8.0 mCi strength. The half-life of
2760Co_{27}^{60}{Co}  is 5.3 years.

Ans:

The strength of the radioactive source is given as:

dNdt=8.0mCi\frac{dN}{dt} = 8.0 mCi

= 8 x 10-3 x 3.7 x 1010

= 29.6 x 107 decay/s

Where,

N = Required number of atoms

Half-life of 2760Co_{ 27 }^{ 60 }{ Co } T 1/2 = 5.3 years

= 5.3 × 365 × 24 × 60 × 60

= 1.67 × 108 s

The rate of decay for decay constant λ is:

dNdt=λN\frac{dN}{dt} = \lambda N

Where, λ=0.693T12=0.6931.67×108s1\lambda = \frac{0.693}{T_\frac{1}{2}} = \frac{0.693}{1.67 \times 10^{8}}s^{-1}

 

N=1λdNdtN = \frac{1}{\lambda} \frac{dN}{dt}

 

=29.6×1070.6931.67×108= \frac{29.6 \times 10^{ 7}}{\frac{ 0.693 }{ 1.67 \times 10^{ 8 }}}

= 7.133 x 1016 atoms

For 2760Co_{27}^{60}{Co} :

Mass of 6.023 × 1023 (Avogadro’s number) atoms = 60 g

Mass of 7.133 x 1016 atoms = =60×7.133×10166.023×1023= \frac{60\times7.133\times10^{16}}{6.023\times10^{23}}

= 7.106 x 10-6 g

Hence, the amount of 2760Co_{27}^{60}{Co} necessary for the purpose is 7.106 × 10−6 g.

Q 13.10 : The half-life of 3890Sr_{ 38 }^{ 90 }{ Sr } is 28 years. What is the disintegration rate of 15 mg of this isotope?

Ans:

Half life of 3890Sr_{ 38 }^{ 90 }{ Sr }, t12t_{\frac{1}{2}} = 28 years

= 28 × 365 × 24 × 60 × 60

= 8.83 × 108 s

Mass of the isotope, m = 15 mg

90 g of 3890Sr_{ 38 }^{ 90 }{ Sr } atom contains 6.023 × 1023 (Avogadro’s number) atoms.

Therefore, 15 mg of 3890Sr_{ 38 }^{ 90 }{ Sr } contains atoms:

= 6.023×1023×15×10390\frac{6.023 \times 10^{23} \times 15 \times 10^{-3}}{90}

i.e 1.0038 x 1020 number of atoms

Rate of disintegration, dNdt=λN\frac{dN}{dt}= \lambda N

Where,

λ\lambda = decay constant = 0.6938.83×108s1\frac{0.693}{8.83 \times 10^{8}}s^{-1}

 

dNdt=0.693×1.0038×10208.83×108\frac{dN}{dt}=\frac{0.693\times 1.0038 \times 10^{20}}{8.83 \times 10^{8}}

= 7.878 x 1010 atoms/s

Hence, the disintegration rate of 15 mg of 3890Sr_{ 38 }^{ 90 }{ Sr } is 7.878 × 1010 atoms/s.

 Q 13.11: Obtain approximately the ratio of the nuclear radii of the gold isotope 79197Au_{ 79 }^{ 197 }{Au} and the 47107Ag_{ 47 }^{ 107 }{Ag}  silver isotope .

Ans:

Nuclear radius of the gold isotope 79197Au_{ 79 }^{ 197 }{Au} = RAu

Nuclear radius of the silver isotope 47107Ag_{ 47 }^{ 107 }{Ag} = RAg

Mass number of gold, AAu = 197

Mass number of silver, AAg = 107

Following is the relationship of the radii of the two nuclei and their mass number:

RAuRAg=(RAuRAg)13\frac{ R_{ Au }}{ R_{ Ag }}=\left (\frac{ R_{ Au }}{ R_{ Ag }} \right )^{\frac{ 1 }{ 3 } } =(197107)13=\left (\frac{ 197 }{ 107 } \right )^{\frac{ 1 }{ 3 } }

= 1.2256

Hence, 1.23 is the ratio of the nuclear radii of the gold and silver isotopes.

 Q 13.12: Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of

(a) 88226Ra_{88}^{226}{Ra}

(b) 86220Rn_{86}^{220}{Rn}

Given

m(88226Ra)m ( _{ 88 }^{ 226 }{ Ra })= 226.02540 u, m(86222Rn)m ( _{ 86 }^{ 222 }{ Rn }) = 222.01750 u,

m(86220Rn)m ( _{ 86 }^{ 220 }{ Rn })= 220.01137 u, m(84216Po)m ( _{ 84 }^{ 216 }{ Po }) = 216.00189 u.

Answer

(a) Alpha particle decay of 88226Ra_{88}^{226}{Ra} emits a helium nucleus. As a result, its mass number reduces to 222 (226 − 4) and its atomic number reduces to 86 (88 − 2). This is shown in the following nuclear reaction.

88226Ra86222Ra+24He_{ 88 }^{ 226 }{ Ra }\rightarrow _{ 86 }^{ 222 }{ Ra }+_{ 2 }^{ 4 }{ He }

Q-value of emitted α-particle = (Sum of initial mass − Sum of final mass) c2

Where, c = Speed of light

It is given that :

m(88226Ra)m ( _{ 88 }^{ 226 }{ Ra })= 226.02540 u

m(86222Rn)m ( _{ 86 }^{ 222 }{ Rn }) = 222.01750 u

m(24He)m ( _{ 2 }^{ 4 }{ He }) = 4.002603 u

Q-value = [226.02540 − (222.01750 + 4.002603)] u c2

= 0.005297 u c2

But 1 u = 931.5 MeV/c2

Q = 0.005297 × 931.5 ≈ 4.94 MeV

Kinetic energy of the α-particle = (Mass  number  after  decayMass  number  before  decay)×Q\left ( \frac{Mass \; number \; after \; decay }{Mass \; number \; before \; decay } \right ) \times Q =222226×4.94= \frac{222}{226} \times 4.94 = 4.85 MeV

(b) Alpha particle decay of 86220Rn_{86}^{220}{Rn}

 

86220Ra84216Po+e4He_{86}^{220}Ra\rightarrow _{84}^{216}Po+_{e}^{4}He

It is given that:

Mass of 86220Rn_{86}^{220}{Rn} = 220.01137 u

Mass of 84216Po_{84}^{216}{Po} = 216.00189 u

Q-value =[ 220.01137 – ( 216.00189 + 4.00260 ) ] x 931.5

≈ 641 MeV

Kinetic energy of the α-particle = =(2204220)×6.41= \left( \frac{ 220 – 4 }{220} \right ) \times 6.41

= 6.29 MeV

Q 13.13: The radionuclide 11C decays according to

611C511 B+e++v:T1/2=20.3 min{ }_{6}^{11} \mathrm{C} \rightarrow_{5}^{11} \mathrm{~B}+e^{+}+v: \quad T_{1 / 2}=20.3 \mathrm{~min}

The maximum energy of the emitted positron is 0.960 MeV.
Given the mass values:
m (116 C) = 11.011434 u and

m (116B ) = 11.009305 u,
calculate Q and compare it with the maximum energy of the positron emitted.

Ans:

611C511 B+e++v:T1/2=20.3 min{ }_{6}^{11} \mathrm{C} \rightarrow_{5}^{11} \mathrm{~B}+e^{+}+v: \quad T_{1 / 2}=20.3 \mathrm{~min}

Mass defect in the reaction is Δm = m (6C11) – m(5B11) – me

This is given in terms of atomic masses. To express in terms of nuclear mass, we should subtract 6me from carbon, 5me from boron.

Δm = m (6C11) – 6me – (m(5B11) – 5me )- me

= m (6C11) – 6me – m(5B11) + 5me – me

= m (6C11) – m(5B11) – 2 me

Δm = [11.011434 – 11.009305 – 2 x 0.000548] u

= 0.002129 – 0.001096 = 0.001033

Q = Δm x 931 MeV

= 0.001033 x 931

Q= 0.9617 MeV

The Q-factor of the reaction is equal to the maximum energy of the emitted positron.

Q 13.14: The nucleus 1023Ne_{10}^{23}{Ne} decays β\beta ^- by emission. Write down theβ\beta  decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:

m(1023Ne)m(_{10}^{23}{Ne}) = 22.994466 u

m(1123Na)m(_{11}^{23}{Na}) = 22.989770 u.

Ans:

In β\beta ^- emission, the number of protons increases by 1, and one electron and an antineutrino is emitted from the parent nucleus.

β\beta ^- emission from the nucleus 1023Ne_{10}^{23}{Ne}.

1023Ne1123Na+e+vˉ+Q_{ 10 }^{ 23 }{ Ne }\rightarrow _{ 11 }^{ 23 }{ Na } + e^{ – } + \bar{ v } + Q

It is given that:

Atomic mass of m(1023Ne)m(_{10}^{23}{Ne})= 22.994466 u

Atomic mass of m(1123Na)m(_{11}^{23}{Na}) = 22.989770 u

Mass of an electron, me = 0.000548 u

Q-value of the given reaction is given as:

Q=[m(1023Ne)[m(1123Na)+me]]c2Q = [m(_{ 10 }^{ 23 }{Ne}) – [m(_{ 11 }^{ 23 }{Na}) + m_e]]c^{ 2 }

There are 10 electrons and 11 electrons in 1023Ne_{ 10 }^{ 23 }{Ne} and 1123Na_{ 11 }^{ 23 }{Na} respectively. Hence, the mass of the electron is cancelled in the Q-value equation.

Q  = [ 22.994466 – 22.989770 ] c2

= (0.004696 c2) u

But 1 u = 931.5 MeV/c2

Q = 0.004696 x 931.5 = 4.374 MeV

The daughter nucleus is too heavy as compared to e and vˉ\bar{v}. Hence, it carries negligible energy. The kinetic energy of the antineutrino is nearly zero. Hence, the maximum kinetic energy of the emitted electrons is almost equal to the Q-value, i.e., 4.374 MeV.

 Q 13.15: 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+13H12H+12H_{ 1 }^{ 1 }{H} + _{ 1 }^{ 3 }{H}\rightarrow _{ 1 }^{ 2 }{ H } + _{ 1 }^{ 2 }{H}

(ii) 612C+612C1020Ne+24He_{ 6 }^{ 12 }{C} + _{ 6 }^{ 12 }{C}\rightarrow _{ 10 }^{ 20 }{ Ne } + _{ 2 }^{ 4 }{He}

Atomic masses are given to be

m(12H)=  2.014102  um( _{ 1 }^{ 2 }{H} ) = \; 2.014102 \; u

m(13H)=  3.016049  um( _{ 1 }^{ 3 }{H} ) = \; 3.016049 \; u

m(612C)=  12.000000  um( _{ 6 }^{ 12 }{C} ) = \; 12.000000 \; u

m(1020Ne)=  19.992439  um( _{ 10 }^{ 20 }{Ne} ) = \; 19.992439 \; u

Ans

(i) The given nuclear reaction is:

11H+13H12H+12H_{ 1 }^{ 1 }{H} + _{ 1 }^{ 3 }{H}\rightarrow _{ 1 }^{ 2 }{ H } + _{ 1 }^{ 2 }{H}

It is given that:

Atomic mass m(11H)=  1.007825  um( _{ 1 }^{ 1 }{H} ) = \; 1.007825 \; u

Atomic mass m(13H)=  3.016049  um( _{ 1 }^{ 3 }{H} ) = \; 3.016049 \; u

Atomic mass m(12H)=  2.014102  um( _{ 1 }^{ 2 }{H} ) = \; 2.014102 \; u

According to the question, the Q-value of the reaction can be written as :

Q = [m(11H)m( _{ 1 }^{ 1 }{H} ) + m(13H)m( _{ 1 }^{ 3 }{H} ) – 2mm(12H)m( _{ 1 }^{ 2 }{H} ) ] c2

= [ 1.007825 + 3.016049 – 2 x 2.014102] c2

Q = ( – 0.00433 c2) u

But 1 u = 931.5 MeV/c2

Q = -0.00433 x 931.5 = – 4.0334 MeV

The negative Q-value of the reaction shows that the reaction is endothermic.

(ii) The given nuclear reaction is:

612C+612C1020Ne+24He_{ 6 }^{ 12 }{C} + _{ 6 }^{ 12 }{C}\rightarrow _{ 10 }^{ 20 }{ Ne } + _{ 2 }^{ 4 }{He}

It is given that:

Atomic mass of m(612C)=  12.000000  um( _{ 6 }^{ 12 }{C} ) = \; 12.000000 \; u

Atomic mass of m(1020Ne)=  19.992439  um( _{ 10 }^{ 20 }{Ne} ) = \; 19.992439 \; u

Atomic mass of m(24He)=  4.002603  um( _{ 2 }^{ 4 }{He} ) = \; 4.002603 \; u

The Q-value of this reaction is given as :

Q = [ 2m(612C)2m( _{ 6 }^{ 12 }{C} ) m(1020Ne)m( _{ 10 }^{ 20 }{Ne} )m(24He)m( _{ 2 }^{ 4 }{He} ) ] c2

= [ 2 x 12.000000 – 19.992439 – 4.002603 ] c2

= [ 0.004958 c2] u

= 0.004958 x 931.5 = 4.618377 MeV

Since we obtained positive Q-value, it can be concluded that the reaction is exothermic.

 Q 13.16: Suppose, we think of fission of a2656Fe_{ 26 }^{ 56 }{Fe} nucleus into two equal fragments, 1328Al_{ 13 }^{ 28 }{Al} . Is the fission energetically possible? Argue by working out Q of the process. Given

m(2656Fe)=  55.93494  um( _{ 26 }^{ 56 }{Fe} ) = \; 55.93494 \; u

m(1328Al)=  27.98191  um( _{ 13 }^{ 28 }{Al} ) = \; 27.98191 \; u

Ans:

The fission of 2656Fe_{ 26 }^{ 56 }{Fe} can be given as :

2656Fe_{ 26 }^{ 56 }{Fe} -> 2 1328Al_{ 13 }^{ 28 }{Al}

It is given that:

Atomic mass of m(2656Fe)=  55.93494  um( _{ 26 }^{ 56 }{Fe} ) = \; 55.93494 \; u

Atomic mass of m(1328Al)=  27.98191  um( _{ 13 }^{ 28 }{Al} ) = \; 27.98191 \; u

The Q-value of this nuclear reaction is given as :

Q = [m(2656Fe)2m(1328Al)m( _{ 26 }^{ 56 }{Fe} ) – 2m( _{ 13 }^{ 28 }{Al} ) ] c2

= [ 55.93494 – 2 x 27.98191 ]c2

= ( -0.02888 c2) u

But 1 u = 931.5 MeV/c2

Q = – 0.02888 x 931.5 = -26.902 MeV

Since the Q-value is negative for the fission, it is energetically not possible.

Q 13.17: The fission properties94239Pu_{ 94 }^{ 239 }{ Pu } of are very similar to those of92235U_{ 92 }^{ 235 }{ U } .The average energy released per fission is 180 MeV. How much energy is released if all the atoms in 1 kg of pure 94239Pu_{ 94 }^{ 239 }{ Pu } undergo fission ?

Ans:

Average energy released per fission of 94239Pu_{ 94 }^{ 239 }{ Pu } , Eav  = 180MeV

Amount of pure 94239Pu_{ 94 }^{ 239 }{ Pu }, m = 1 kg = 1000 g

NA= Avogadro number = 6.023 × 1023

Mass number of 94239Pu_{ 94 }^{ 239 }{ Pu } = 239 g

1 mole of 94239Pu_{ 94 }^{ 239 }{ Pu } contains NA atoms.

Therefore, mg of 94239Pu_{ 94 }^{ 239 }{ Pu } contains (NAMass  Number×m)\left ( \frac{N_A}{Mass \; Number}\times m \right ) atoms

6.023×1023239×1000=2.52×1024\frac{6.023 \times 10^{23}}{239}\times 1000 = 2.52 \times 10^{ 24 } atoms

Total energy released during the fission of 1 kg of 94239Pu_{ 94 }^{ 239 }{ Pu } is calculated as:

E = Eav x 2.52 x 1024

= 180 x 2.52 x 1024

= 4.536 x 1026 MeV

Hence, 4.536 x 1026 MeV is released if all the atoms in 1 kg of pure 94239Pu_{ 94 }^{ 239 }{ Pu } undergo fission.

Q 13.18: A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much  92235 U did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of  92235U and that this nuclide is consumed only by the fission process.

Ans:

Reactor consumes half its fuel is 5 years. Therefore, the half-life of the fuel of the fission reactor, t1/2 = 5 x 365 x 24 x 60 x 60 s

200 MeV is released during the fission of 1 g of 92235U.

1 mole, i.e., 235 g of 92235U contains 6.023 x 1023 atoms.

Number of atoms in 1 g of 92235U  is (6.023 x 1023/235) atoms.

The total energy generated per gram of 92235U  is (6.023 x 1023/235) x 200 MeV/g

=200×6.023×1023×1.6×1019×106235= \frac{200 \times 6.023 \times 10^{23}\times 1.6\times 10^{-19}\times 10^{6}}{235}

= 8.20 x 1010 J/g

The reactor operates only 80% of the time.

Therefore, the amount of 92235U  consumed in 5 years by the 1000 MW fission reactor is

=5×365×24×60×60×1000×1068.20×1010×80100= \frac{5\times 365\times24\times 60\times 60 \times 1000\times 10^{6} }{8.20\times 10^{10}}\times \frac{80}{100}

= 12614400/8.20

= 1538341 g

= 1538 Kg

Initial amount of 92235U is 2 x 1538 = 3076 Kg

Q 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

12H+12H23He+n+3.27MeV{ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \rightarrow{ }_{2}^{3} \mathrm{He}+\mathrm{n}+3.27 \mathrm{MeV}

Ans:

The fusion reaction given is

12H+12H23He+n+3.27MeV{ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \rightarrow{ }_{2}^{3} \mathrm{He}+\mathrm{n}+3.27 \mathrm{MeV}

Amount of deuterium, m = 2 kg

1 mole, i.e., 2 g of deuterium contains 6.023 x 1023 atoms.

Therefore, 2 Kg of deuterium contains (6.023 x 1023/2) x 2000 = 6.023 x 1026 atoms

From the reaction given, it can be understood that 2 atoms of deuterium fuse, 3.27 MeV energy are released. The total energy per nucleus released during the fusion reaction is

E = (3.27/2) x 6.023 x 1026  MeV

= (3.27/2) x 6.023 x 1026  x 1.6 x 10 -19 x 106

= 1.576 x 1014 J

Power of the electric lamp, P = 100 W = 100 J/s

The energy consumed by the lamp per second = 100 J

The total time for which the electric lamp will glow is 1.576 x 1014 /100 = 1.576 x 1012s

= (1.576 x 1012)/ (365 x 24 x 60 x 60)

= (1.576 x 1012)/3.1536 x107

= 4.99 x 104 years

Q 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.)

Ans :

When two deuterons collide head-on, the distance between their centers, d is given as:

Radius of 1st deuteron + Radius of 2nd deuteron

Radius of a deuteron nucleus = 2 f m = 2 × 10−15 m

d = 2 × 10−15 + 2 × 10−15 = 4 × 10−15 m

Charge on a deuteron nucleus = Charge on an electron = e = 1.6 × 10−19 C

Potential energy of the two-deuteron system:

v=e24πϵ0dv = \frac{ e^{ 2 } }{ 4 \pi \epsilon_0 d }

Where,

ϵ0\epsilon_0= Permittivity of free space

14πϵ0=9×109Nm2C2\frac{1}{4 \pi \epsilon_0} = 9 \times 10^{9} Nm^2C^{-2}

Therefore,

V=9×109×(1.6×1019)24×1015JV = \frac{9 \times 10^{9} \times (1.6 \times 10^{-19})^2}{4 \times 10^{-15} } J

 

V=9×109×(1.6×1019)24×1015×(1.6×1019)eVV = \frac{9 \times 10^{9} \times (1.6 \times 10^{-19})^2}{4 \times 10^{-15} \times (1.6 \times 10^{-19})}eV

= 360 k eV

Hence, the height of the potential barrier of the two-deuteron system is 360 k eV.

Q 13.21: From the relation R = R0A1/3, where R0 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).

Ans:

Nuclear radius is given as

R = R0A1/3

Here,

R0 is Constant

A is the mass number of the nucleus

Nuclear matter density, ρ = Mass of the nucleus/Volume of the nucleus

Mass of the nucleus = mA

Density of the nucleus = (4/3)πR3

= (4/3)π(R0A1/3)3

= (4/3)πR03A

ρ = mA/[(4/3)πR03A]

= 3mA/(4πR03A)

ρ = 3m/(4πR03)

Nuclear matter density is nearly a constant and is independent of A.

Q 13.22: For the β+\beta^{+} (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++ZAXZ1AY+ve^+ + _{Z}^{A}{X} \rightarrow _{ Z-1 }^{A}{Y}+v

 Show that ifβ+\beta^{+} emission is energetically allowed, electron capture is necessarily allowed but not vice−versa.

Ans :

The chemical equations of the two competing process are written as follows:

AZXAZ1Y+e+ν+Q1_{A}^{Z}\textrm{X}\rightarrow _{A}^{Z-1}\textrm{Y}+e+\nu+Q_1 [Positron Emission]

e1+AZXAZ1Y+v+Q2e^{-1}+_{A}^{Z}\textrm{X}\rightarrow _{A}^{Z-1}\textrm{Y}+v+Q_2 [Electron Capture]

The Q-value of the positron emission reaction is determined as follows :

Q1=[mN(ZAX)mN(Z1AY)me]c2Q1=[m(ZAX)Zmem(Z1AY)+(Z1)meme]c2Q1=[m(ZAX)m(Z1AY)2me]c2\\Q_{1}=[m_{N}(_{Z}^{A}X)-m_{N}(_{Z-1}^{A}Y)-m_{e}]c^{2} \\ \\Q_{1}=[m(_{Z}^{A}X)-Zm_{e}-m(_{Z-1}^{A}Y)+(Z-1)m_{e}-m_{e}]c^{2} \\ \\Q_{1}=[m(_{Z}^{A}X)-m(_{Z-1}^{A}Y)-2m_{e}]c^{2}

The Q-value of the electron capture reaction is determined as follows:

Q2=[mN(ZAX)+memN(Z1AY)]c2Q2=[m(ZAX)m(Z1AY)]c2\\Q_{2}=[m_{N}(_{Z}^{A}X)+m_{e}-m_{N}(_{Z-1}^{A}Y)]c^{2} \\ \\Q_{2}=[m(_{Z}^{A}X)-m(_{Z-1}^{A}Y)]c^{2}

In the equation, mis the mass of nucleus and m denotes the mass of atom.

From equation (1) and (2), we understand that,

Q1 = Q2 – 2mec2

From the above relation, we can infer that if Q1 > 0, then Q2 > 0; Also, if Q2> 0, it does not necessarily mean that Q1 > 0.

In other words, this means that if  emission is energetically allowed, then the electron capture process is necessarily allowed, but not vice-versa. This is because the Q-value must be positive for an energetically allowed nuclear reaction.

Q 13.23: In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are 2412Mg (23.98504u), 2512Mg(24.98584u) and 2612Mg (25.98259u). The natural abundance of 2412Mg is 78.99% by mass. Calculate the abundances of the other two isotopes.

Ans:

Let the abundance of 2512Mg be x%

The abundance of 2612Mg = (100 – 78.99 -x)%

= (21.01 – x)%

The average atomic mass of magnetic

24.312=23.98504×78.99+24.98584x+25.98259(21.01x))10024.312 = \frac{23.98504\times 78.99+24.98584x+25.98259(21.01-x))}{100}

24.312  x 100= 1894.57 + 24.98584x + 545. 894 – 25.98259x

2431.2 = 2440.46 – 0.99675x

0.99675x = 2440.46 – 2431.2

0.99675x =  9.26

x = 9.26/0.99675

x = 9.290%

Abundance of 2512Mg is 9.290%

The abundance of 2612Mg = (21.01 – x)%

= (21.01 -9.290 )% = 11.71%

Q 13. 24: The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei 4120Ca and 2713Al from the following data:
m(4020Ca ) = 39.962591 u
m(4120Ca ) = 40.962278 u
m(2613Al ) = 25.986895 u
m(2713Al ) = 26.981541 u

Ans:

When the nucleons are separated from 20Ca41 → 20 Ca40 + 0n1

Mass defect, Δm = m (20 Ca40) + mn – m(20 Ca41)

= 39.962591 + 1.008665 – 40.962278

= 0.008978 amu

Neutron separation energy = 0.008978  x 931 MeV = 8.362 MeV

Similarly, 13Al27 → 13 Al26 + 0n1

Mass defect, Δm = m (13 Al26) + mn – m(13Al27)

= 25.986895 + 1.008665 – 26.981541

= 26.99556 – 26.981541 = 0.014019 amu

Neutron separation energy = 0.014019 x 931 MeV

= 13.051 MeV

Q 13. 25: A source contains two phosphorous radionuclides 3215P (T1/2 = 14.3d) and 3315P(T1/2 = 25.3d). Initially, 10% of the decays come from 3315 P. How long one must wait until 90% do so?

Ans:

Rate of disintegration is

– (dN/dt) ∝ N

Initially 10% of the decay is due to 3315 P and 90% of the decay comes from 3215P.

We have to find the time at which 90% of the decay is due to 3315 P and 10% of the decay comes from 3215P.

Initially, if the amount of 3315 P is N, then the amount of 3215P is 9N.

Finally, if the amount of 3315 P is 9N’, then the amount of 3215P is N’.

For 3215P,

N9N=(12)t/T1/2\frac{N’}{9N}=\left ( \frac{1}{2} \right )^{t/T_{1/2}}

 

N=9N(12)t/T1/2N’=9N \left ( \frac{1}{2} \right )^{t/T_{1/2}}

 

N=9N(2)t/14.3N’=9N \left ( 2\right )^{-t/14.3}—–(1)

For 3315 P

9NN=(12)t/T1/2\frac{9N’}{N}=\left ( \frac{1}{2} \right )^{t/T_{1/2}}

 

9N=N(12)t/T1/29N’= N\left ( \frac{1}{2} \right )^{t/T_{1/2}}

 

9N=N(2)t/25.39N’= N\left ( 2 \right )^{-t/25.3}——(2)

On dividing, (1) by (2)

19=9×2t25.3t14.3181=211t25.3×14.3log1log81=11t25.3×14.3log211t25.3×14.3=01.9080.301t=25.3×14.3×1.90811×0.301208.5days\\\frac{1}{9}=9\times 2^{\frac{t}{25.3}-\frac{t}{14.3}} \\ \\\frac{1}{81}=2^{-\frac{11t}{25.3\times 14.3}} \\ \\log1-log81=-\frac{11t}{25.3\times 14.3}log2 \\ \\-\frac{11t}{25.3\times 14.3}=\frac{0-1.908}{0.301} \\ \\t=\frac{25.3\times 14.3\times 1.908}{11\times 0.301}\approx 208.5 \: days

Hence, it will take about 208.5 days for 90% decay of 15P33.

Q 13. 26: Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle. Consider the following decay processes:

88223Ra82209 Pb+614C{ }_{88}^{223} \mathrm{Ra} \rightarrow{ }_{82}^{209} \mathrm{~Pb}+{ }_{6}^{14} \mathrm{C}

88223Ra86219Rn+24He{ }_{88}^{223} \mathrm{Ra} \rightarrow{ }_{86}^{219} \mathrm{Rn}+{ }_{2}^{4} \mathrm{He}

Calculate the Q-values for these decays and determine that both are energetically allowed.

Ans:

88223Ra82209 Pb+614C{ }_{88}^{223} \mathrm{Ra} \rightarrow{ }_{82}^{209} \mathrm{~Pb}+{ }_{6}^{14} \mathrm{C}

Δm = m (88Ra223) – m (82Pb209) – m(6C14)

= 223.01850 – 208.9817 – 14.00324

= 0.03419 u

The amount of energy released is given as

Q = Δm x 931 MeV

= 0.03419 x 931 MeV = 31.83 MeV

88223Ra86219Rn+24He{ }_{88}^{223} \mathrm{Ra} \rightarrow{ }_{86}^{219} \mathrm{Rn}+{ }_{2}^{4} \mathrm{He}

Δm = m (88Ra223) – m (86Rn219) – m(2He4)

= 223.01850 – 219.00948 – 4.00260

= 0.00642 u

Q = 0.00642 x 931 MeV = 5.98 MeV

As both the Q-factor values are positive the reaction is energetically allowed.

Q 13. 27: Consider the fission of 23892U by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are 14058Ce and 9944Ru. Calculate Q for this fission process. The relevant atomic and particle masses are

m(23892U ) =238.05079 u
m(14058Ce ) =139.90543 u
m(9944Ru ) = 98.90594 u

Ans:

In the fission of 23892U, 10 β particles decay from the parent nucleus. The nuclear reaction can be written as:

92238U+01n58140Ce+4499Ru+1010e{ }_{92}^{238} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{58}^{140} \mathrm{Ce}+{ }_{44}^{99} \mathrm{Ru}+10{ }_{-1}^{0} e

Given:

m1 = (23892U ) =238.05079 u
m2 = (14058Ce ) =139.90543 u
m3 = (9944Ru ) = 98.90594 u

m4 = 10n  = 1.008665 u

The Q value is given as

Q = [ m'(23892U ) + m (10n) – m'(14058Ce ) – m'(9944Ru ) – 10me ] c2

Here,

m’ is the atomic masses of the nuclei

m'(23892U ) = m1 – 92 me

m'(14058Ce ) = m2 – 58me

m'(9944Ru ) = m3 – 44me

Therefore, Q = [m1 – 92 me+ m4 – m2 + 58m– m3 + 44me– 10me ] c2

= [m1 + m4 – m2 – m3]c2

= [238.0507 + 1.008665 – 139.90543 – 98.90594]c2

= 0.247995 c u

[1 u = 931.5 MeV/c2]

Q = 0.247995  x 931.5 = 231.007 MeV

The Q- value of the fission process is 231.007 MeV

Q 13. 28: Consider the D–T reaction (deuterium–tritium fusion)
12H+13H24He+n{ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He}+\mathrm{n}

(a) Calculate the energy released in MeV in this reaction from the
data:
m(21H )=2.014102 u
m(31H ) =3.016049 u
(b) Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the reaction? (Hint: Kinetic energy required for one fusion event =average thermal kinetic energy available with the interacting particles = 2(3kT/2); k = Boltzman’s constant, T = absolute temperature.)

Ans:

12H+13H24He+n{ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He}+\mathrm{n}

The Q value is given as

Q = Δm x 931 MeV

= (m (1H2) + m(1H3) – m(2He4) – mn) x 931

=( 2.014102 + 3.016049- 4.002603-1.00867) x 931

Q = 0.0188 x 931 = 17.58 MeV

(b) Repulsive potential energy of two nuclei when they touch each other is

=e24πϵ0(2r)=\frac{e^{2}}{4\pi \epsilon _{0}(2r)}

Where ε0 = Permittivity of free space

14Πϵ0\frac{1}{4\Pi \epsilon _{0}}=9 x 109 Nm2C-2

=9×109×(1.6×1019)22×2×1015joule= \frac{9\times 10^{9}\times (1.6\times 10^{-19})^{2}}{2\times 2\times 10^{-15}} joule