Nuclei Class 12th Physics Part Ii CBSE Solution

Class 12th Physics Part Ii CBSE Solution
Exercises
  1. Two stable isotopes of lithium^6 3Li and^7 3Li have respective abundances of 7.5% and…
  2. Boron has two stable isotopes,^10 5B and^11 5B. Their respective masses are 10.01294 u and…
  3. The three stable isotopes of neon: 2010Ne,^21 10Ne and^22 10Ne, have respective abundances…
  4. Obtain the binding energy (in MeV) of a nitrogen nucleus (^14 7N) given m (^14 7N) =…
  5. Obtain the binding energy of the nuclei^56 26Fe and^209 83BI in units of MeV from the…
  6. A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to…
  7. Write nuclear reaction equations for (i) α-decay of^226 88Ra (ii) α-decay of^242 94Pu…
  8. A radioactive isotope has a half-life of T years. How long will it take the activity to…
  9. The normal activity of living carbon-containing matter is found to be about 15 decays per…
  10. Obtain the amount of^60 27Co necessary to provide a radioactive source of 8.0 mCi…
  11. The half-life of^90 38Sr is 28 years. What is the disintegration rate of 15 mg of this…
  12. Obtain approximately the ratio of the nuclear radii of the gold isotope^197 79Au and the…
  13. Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of…
  14. The radionuclide^11 C decays according to The maximum energy of the emitted positron is…
  15. The nucleus^23 10Ne decays by β- emission. Write down the β-decay equation and determine…
  16. The Q value of a nuclear reaction A + b → C + d is defined by Q = [mA + mb- mC - md]c^2…
  17. Suppose, we think of fission of a^56 26Fe nucleus into two equal fragments,^28 13Al. Is…
  18. The fission properties of^239 94Pu are very similar to those of^235 92U. The average…
  19. A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much^235 92U did it…
  20. How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium?…
  21. Calculate the height of the potential barrier for a head-on collision of two deuterons.…
  22. From the relation R = R0A1/3, where R0 is a constant and A is the mass number of a…
  23. For the β + (positron) emission from a nucleus, there is another competing process known…
Additional Exercises
  1. In a periodic table, the average atomic mass of magnesium is given as 24.312 u. The…
  2. The neutron separation energy is defined as the energy required to remove a neutron from…
  3. A source contains two phosphorous radio nuclides^32 15P (T1/2 = 14.3d) and^33 15P (T1/2 =…
  4. Under certain circumstances, a nucleus can decay by emitting a particle more massive than…
  5. Consider the fission by fast neutrons. In one fission event, no neutrons are emitted and…
  6. Consider the D-T reaction (deuterium-tritium fusion) (a) Calculate the energy released in…
  7. Obtain the maximum kinetic energy of β-particles, and the radiation frequencies of γ decay…
  8. Calculate and compare the energy released by a) fusion of 1.0 kg of hydrogen deep within…
  9. Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten…

Exercises
Question 1.

Two stable isotopes of lithium 63Li and 73Li 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.


Answer:

The atomic mass of the two isotopes are respectively given as 6.01512 u and 7.01600 u with have respective abundances of 7.5% and 92.5%.

Hence the mass of Lithium can be given as,


MLi = 


 = 6.9409u


The mass of Lithium is = 6.9409 u



Question 2.

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:

The atomic mass of the two isotopes of Boron (105B and 115B) are given as 10.01294 u and 11.00931 u


Let their abundances be respectively, x and (100-x) percent.


Hence from the formula discussed in previous question,


10.811 = 


So, x = 19.89% and (100-x) = 80.11%


Hence the abundance of 10B5 is 19.89% and that of 11B5 is 80.11%



Question 3.

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 the three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon.


Answer:

The atomic mass and abundance of 20Ne10 is m1 = 19.99 u, n1 = 90.51%

The atomic mass and abundance of 21Ne10 is m2 = 20.99 u, n2 = 0.27%


The atomic mass and abundance of 22Ne10 is m3 = 21.99 u, n3 = 9.22%


Mass = = 


Hence average atomic mass of Neon = 


 u


= 20.1771 u


Average atomic mass of Neon = 20.1771u



Question 4.

Obtain the binding energy (in MeV) of a nitrogen nucleus (147N) given m (147N) = 14.00307 u


Answer:

The number of neutron and Proton both are 7 in a 14N7 atom.

The mass defect Δm = mp + mn-mN


And Binding energy is given by EB = Δmc2


Mass of a proton = 1.007825 u


Mass of a neutron = 1.008665 u


In this case mass defect = Δm = 7 × (1.007825 + 1.008665)-14.00307u = 0.11236 u


We know, 1 u = 931.5 MeV/c2


So, E = Δmc2 = 0.11236 × 931.5 (MeV/c2 × c2 ) = 104.66334 MeV


Hence, Binding energy of 14N7 = 104.66334 MeV



Question 5.

Obtain the binding energy of the nuclei 5626Fe and 20983BI in units of MeV from the following data:

m (5626Fe) = 55.934939 u m (20983BI) = 208.980388 u


Answer:

The number of neutron and Proton both in a 56Fe26 atom is respectively 30 and 26.

The mass defect Δm = mp + mn-mFe


And Binding energy is given by EB = Δmc2


Mass of a proton = 1.007825 u


Mass of a neutron = 1.008665 u


In this case mass defect = Δm = (26 × 1.007825 + 30 × 1.008665-55.934939)u = 0.528461u


We know, 1 u = 931.5 MeV/c2


So, E = Δmc2 = 0.528461 × 931.5(MeV/c2) × c2 = 492.26MeV


So, Binding Energy per Nucleon =  =  = 8.79MeV


The number of neutron and Proton both in a 209Bi83 atom is respectively 126 and 83.


The mass defect Δm = mp + mn-mBi


And Binding energy is given by EB = Δmc2


Mass of a proton = 1.007825 u


Mass of a neutron = 1.008665 u


Mass Defect = Δm = 83 × 1.007825 + 126 × 1.008665-208.980388 = 1.760877 u


We know, 1 u = 931.5 MeV/c2


So, E = Δmc2 = 1.760877 × 931.5 (Mev/c2) × c2 = 1640.26 MeV


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



Question 6.

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 a copper coin = 3 g


Atomic mass of copper atom, m = 62.92960u


The total number of atoms in the coin = N = 


Where,


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


Hence, N =  = 2.868 × 1022


The number of neutron and Proton both in a 63Cu29 atom is respectively 34 and 29.


The mass defect Δm = mp + mn-mFe


And Binding energy is given by EB = Δmc2


Mass of a proton = 1.007825 u


Mass of a neutron = 1.008665 u


Mass defect = Δm = 29 × 1.007825 + 34 × 1.008665-62.92960 u = 0.591935u


Mass defect of the coin = Δm = 0.591935 × 2.868 × 1022 u = 1.69766958 × 1022 × 931.5 MeV/c2


Binding Energy = 1.581 × 1025MeV = 1.581 × 1025 × 1.6 × 10-13 J = 2.5296 × 1012 J


Hence, 2.5296 × 1012 Joules of energy is required to separate the nucleons of the coin.



Question 7.

Write nuclear reaction equations for

(i) α-decay of 22688Ra

(ii) α-decay of 24294Pu

(iii) β-decay of 3215P

(iv) β-decay of 21083BI

(v) β+ -decay of 116C

vi) β+ -decay of 4397Tc

(vii) Electron capture of 12054Xe


Answer:

Explanation for the below-given solution in a nutshell, In case of α decay, there is a loss of 2 protons and 4 neutrons, In case of β + decay there is a loss of a proton and a neutrino is emitted, and for every β- decay there is a gain of one proton and an antineutrino is emitted.


Hence the equations for the given cases will be:


i)  (α-decay)


ii)  (α-decay)


iii)  (β-decay)


iv)  (β-decay)


v)  (β + -decay)


vi)  (β + -decay)


vii)  (electron capture)



Question 8.

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:

Suppose, Initially the amount of radioactive isotope is N0

After time t if x% of it’s original value remains, and let λ be the decay constant


Then, we can write 


∴ λt = ln[100/x]


we know that, ∴ 


Hence, 


Given that 


a) If x = 3.125% then t =  = ≈ 5T years


b) If x = 1% then, t =  = 6.645T years



Question 9.

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 radioactive 146C present with the stable carbon isotope 126C. 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) of 146C, and the measured activity, the age of the specimen can be approximately 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 civilization.


Answer:

Let N be a number of radioactive carbon found in normal carbon and N0 be the number of radioactive carbon found in the specimen. The half-life of C-14 is 5730 yrs.

Decay rate of living carbon-containing matter = D = 15 decay/min-gm


Decay rate of the specimen at Mohenjo-Daro = D0 = 9 decays/min-gm


From the exponential decay rate law, we get,


∴ Hence, 


So, -λt = ln(3/5) = -0.5108


t = 


or, t = 


Approximate age of Indus-Valley-Civilization is 4223.5 years.



Question 10.

Obtain the amount of 6027Co necessary to provide a radioactive source of 8.0 mCi strength. The half-life of 6027Co is 5.3 years.


Answer:

The Strength a radioactive element is given by = 

And, 1Ci = 3.7 × 1010decays/sec;


∵ hence, 8.0 mCi = 8 × 10-3 × 3.7 × 1010


= 29.6 × 107 decay/sec (∵ 1 Ci = 3.7 × 1010 decay/sec)


Half life of Co60 is 5.3 years


= 5.3 × 365 × 86400 sec (∵ 1day = 86400 sec)


= 1.67 × 108 s-1


We know,


N =  = 


Mass of 7.133 × 1016 atoms will be =  gms


Hence, we need  gms of Co60.



Question 11.

The half-life of 9038Sr is 28 years. What is the disintegration rate of 15 mg of this isotope?


Answer:

From the principle of exponential radioactive decay,

we know that 


Here, the mass of the isotope is = 15 g


No. of atoms in 15gm of atom is =  atoms


And, λ = (0.693/T1/2) where, Half life = 28 × 365 × 86400 = 8.83 × 108 secs


Hence,


The disintegration rate of 15 mg of the isotope will be 



Question 12.

Obtain approximately the ratio of the nuclear radii of the gold isotope 19779Au and the silver isotope.


Answer:

We know that, RA1/3 (where R is nuclear radii and A is the mass number)

∵ Hence, 


So, the ratio of their radii will be


 = 1.2256



Question 13.

Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of (a) 22688Ra and (b) 22086Rn

Given m (22688Ra) = 226.02540 u, m (22286Rn) = 222.01750 u,

m (22286Rn) = 220.01137 u, m (21684Po) = 216.00189 u.


Answer:

a) Q value of emitted α particle = (Total Initial mass-Total final mass) × c2

∴ The α particle decay of 226Ra88 is given by:



Hence the Q value will be = [226.02540-(222.01750 + 4.002603)] × c2


= 0.005297 u × c2


= 0.005297 × 931.5 Mev


= 4.94MeV


Kinetic energy is given by = 


 × 4.94 Mev


= 4.85 MeV


b) Similarly, the decay of 220Rn86 is given by :



Hence, Q value = [220.01137-(216.00189 + 4.0026)]u × c2 = 0.00688u × 931.5 MeV = 6.41MeV


Kinetic energy = 


= 6.41 ×  MeV


= 6.293 MeV



Question 14.

The radionuclide 11C decays according to



The maximum energy of the emitted positron is 0.960 MeV.

Given the mass values:

m (116C) = 11.011434 u and m (115B) = 11.009305 u,

calculate Q and compare it with the maximum energy of the positron emitted.


Answer:

Important: We must consider electron mass in β decays, this mass is no more negligible.


The nuclear reaction is given by :



If atomic masses are used instead of nuclear masses, then we have to add 6 me in the case of 11Cand 5 me in the case of 11B one excess electron, one electron is already there so in the final equation there will be 2 electrons.


Hence Q value for this reaction is given by = [11.011434-(11.009305 + 2 × me)] × c2


We know, me = 0.000548 u


Q = [11.011434-(11.009305 + 2 × 0.000548)] × c2


= 0.001033 u × c2


= 0.962 MeV


Hence the Q value is comparable with the maximum energy of the positron emitted.



Question 15.

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 (2311Ne) = 22.089770 u.


Answer:

For β- decay,the number of proton increases by 1.The reaction is given as:


Here the electron masses gets cancelled as Na has one more electron than Ne hence,


Q = (22.994466-22.089770)u × c2


= 0.00469 × 931.5 MeV


= 4.37 MeV


The maximum k.E of the emitted electrons are comparable to the Q value; Hence the maximum K.E of the electrons emitted will be = 4.37 MeV



Question 16.

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 (2010Ne) = 19.992439 u


Answer:

i) The given equation is: 

Hence, according to given data and given definition of Q value in the question


Q value = [m(1H1) + m(3H1)-2 × m(2H1)] × c2


= (-0.00433u × c2) =


-0.00433 × 931.5 MeV


= -4.0334 MeV


The negative value of Q value suggests this reaction is endothermic.


ii) The reaction is : 


According to previous solution Q value,


Q = [2 × m(12C6)-m(20Ne10)-m(4He2)] × c2


= (0.004958u) × c2


= 0.004958 × 931.5


= 4.6183MeV


Here, the Q value is positive hence reaction is exothermic.



Question 17.

Suppose, we think of fission of a 5626Fe nucleus into two equal fragments, 2813Al. Is the fission energetically possible? Argue by working out Q of the process. Given m (5626Fe) = 55.93494 u and m (2813Al) = 27.98191 u.


Answer:

The fission reaction can be given as : 

The Q value for this reaction will be given as = [m()-2 × m()] × c2


= (55.93494-2 × 27.98191)u × c2 MeV


= -0.02888u × c2 MeV


= -0.02888 × 931.5


= -26.902MeV


The Q value is negative which suggests this reaction is endothermic, but we know fission reactions are exothermic. Hence, this fission reaction is not energetically possible.



Question 18.

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:

The atomic mass of 239Pu94 is 239.

Hence there are 6.023 × 1023 atoms in 239 gm.


So, there are  = 2.52 × 1024 atoms


The energy released will be = average E × No. of atoms


Hence, E = 180 × 2.52 × 1024 Mev = 4.536 × 1026MeV


if all the atoms in 1 kg of pure 239Pu undergo fission then 4.536 × 1026MeV



Question 19.

A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much 23592U 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:

Half-life of the fuel in the fission reactor = 5 years

= 5 × 365 × 86400 s (∵ 1 day = 86400 seconds)


= 1.576 × 108 sec


∵ 1 gm of Uranium fission gives 200 MeV energy [This can be worked out from the fission equation given below of Uranium].



∴ 1 gm of Uranium, = 


Energy generated by Uranium fission = 


 J


= 8.2 × 1010 J/gm


∵ The 1000MW reactor operates only 80% of it’s time,hence in 5 years amount of uranium consumed



= 1538 kg


∴ This amount is consumed within the half life time.


Hence, the initial amount will be = 2 × 1538Kg


= 3076Kg



Question 20.

How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as:

21H + 21H → 32He + n + 3.27 MeV


Answer:

The reaction is given As:


Amount of Deuterium fuel = 2.0 kg


∵ 2 gm of deuterium contains 6.023 × 1023 atoms


Hence, 2 kg of deuterium contains 6.023 × 1023 × 103 atoms


From the reaction we can infer 2 g of deuterium gives 3.27 MeV energy


∴ Hence, Total energy released in this reaction = MeV


= 9.847 × 1026 × 106 × 1.6 × 10-19J


= 1.576 × 1014 J.


∵ The power of the lamp is 100 W = 100J/s


Hence, time of glow by this much energy is = 


= 1.576 × 1012 sec


= (1.576 × 1012)/(60 × 60 × 24)


= 4.9 × 104 years.



Question 21.

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:

In case of collision of two deuterons, the distance between their centres d is given as

∵ d = Radius of first atom + Radius of 2nd atom


radus of deuteron = 2 fm = 2 × 10-15 m


Hence, d = 2 × 2 × 10-15 = 4 × 10-15 m


Potential energy of this two deuteron system will be V = 


where Є0 = permittivity of free space


And  = 9 × 109 m2C-2


J


∴ 


∴ V = 360KeV


Hence, the height of barrier potential is 360 keV.



Question 22.

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).


Answer:

The radius od a nucleus is given as R = R0A1/3 (A = mass number)

And, the density is given by = 


Let m be average mass of the atom, hence Mass = mA


And volume = 


Hence, 𝞺 =  = 


This is a constant independent of A. So, Nuclear density is a constant.



Question 23.

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 reaction for the electron capture process is given by:


The reaction for the β + emission is given by:



Considering the atomic mass of the nucleus we get the following Q value equations.


QA = 



And similarly,


QB = m



Here from these two equations we can see if QB > 0, then QA > 0 But, If QA > 0 that doesn’t imply QB > 0


Which means, if β + emission is energetically allowed, electron capture is necessarily allowed but not vice–versa.




Additional Exercises
Question 1.

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 other two isotopes.


Answer:

Given that,


Average atomic mass of magnesium, mavg = 24.312 u


Mass of magnesium isotope (m1), 


Mass of magnesium isotope (m2), 


Mass of magnesium isotope (m3), 


Abundance of magnesium isotope (n1), 


We know that sum of abundances of all isotopes together is equal to total magnesium available on earth, 100% magnesium.


Let, Abundance of magnesium isotope (n2


Thus, Abundance of isotope (n3), 


We have the relation for the average atomic mass is,



∴ 


⇒ 


⇒ -9.272526 u = -0.99675x u


∴  = Abundance of isotope (n2), 


Thus, Abundance of isotope (n3), 



Question 2.

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


Answer:

Given that,


Mass 


Mass 


Mass 


Mass 


We know that,


Removal of one neutron from leads to the formation of ,



The mass defect of this reaction is,




We have, 1 u = 913.5 MeV/c2


Hence the energy required to remove neutron removal,



We know that,


Removal of one neutron from leads to the formation of ,



The mass defect of this reaction is,




We have, 1 u = 913.5 MeV/c2


Hence the energy required to remove neutron removal,



Hence, Energy required for removal of neutron from,





Question 3.

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


Answer:

Half-life of 


Half-life of 



Thus the source have initially 10% of .


Suppose after some time t, this situation may be reversed.


Let,


Initial number of nucleus of 


Initial number of nucleus of 


For 



∴  ……………(1)


For 



∴  ……………(2)


On dividing equation (1) by equation (2) we get,


⇒ 


⇒ 


⇒ 


⇒ 


∴ 


Thus, 



Question 4.

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



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


Answer:


Where,


Δm = Mass defect (or) mass lost during reaction


c = speed of light


Take nuclear emission reaction given,



Energy (Heat) released during nuclear emission reaction,


We know that,


Mass of , m1 = 223.0185 u


Mass of , m2 = 208.98107 u


Mass of , m3 = 14.00324 u


∴ 


But, 1 u = 931.5 MeV/c2


∴ 


So, this reaction results in the emission of 31.848 MeV of energy.


Take nuclear emission reaction given,



Energy (Heat) released during nuclear emission reaction,


We know that,


Mass of , m1 = 223.0185 u


Mass of , m2 = 219.00948 u


Mass of , m3 = 4.00260 u


∴ 


But, 1 u = 931.5 MeV/c2


∴ 


So, this reaction results in the emission of 5.98 MeV of energy.


Since, both reactions are giving energy outside ( + ve), given reactions are energetically allowed.



Question 5.

Consider the fission 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


Answer:

Given that,


Mass of a nucleus , m1 = 238.05079 u


Mass of a nucleus , m2 = 139.90543 u


Mass of a nucleus , m3 = 98.90594 u


Mass of a neutron , m4 = 1.008665 u


In the fission of , 10 beta particles are emitted. The nuclear reaction is,



Energy released during above nuclear fission is given by,



Where,


m’ represents the corresponding atomic masses of the nuclei


 = m1 – 92me (∵ atomic number 92)


 = m2 – 58me (∵ atomic number 58)


 = m3 – 44me (∵ atomic number 44)


 = m4


∴ 




= (0.247995)u×c2


But, u = 931.5 MeV/c2


∴ Q = 0.247995×931.5 Mev


= 231.007 MeV


The energy released during given fission reaction is 231.007 MeV.



Question 6.

Consider the D–T reaction (deuterium-tritium fusion)



(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 = Boltzmann’s constant, T = absolute temperature.)


Answer:

a) Given,


Mass of , m1 = 2.014102 u


Mass of , m2 = 3.016049 u


Mass of , m2 = 4.002603 u


Mass of , m3 = 1.008665 u



Where,


Δm = Mass defect (or) mass lost during reaction


c = speed of light


Given nuclear fusion reaction is,



∴ 




But, u = 931.5 MeV/c2


∴ 


b) Given,


Radius of the deuterium and tritium, r = 2fm = 2×10-15 m


Charge on deuterium and tritium nuclei = e = 1.6×10-19 C


Thus,


Distance between the two nuclei, d = r + r = 4×10-15 m


Repulsive potential energy between two nuclei is,



Where,


e = charge


ϵ0 = permittivity of the space


d = distance between charges


and, 


∴ 




(∵1.6×10-19 C = 1eV)


Hence, it needs 360 keV of kinetic energy to overcome coulomb repulsion.


But, given that Kinetic energy (KE) is,



Where,


k = Boltzmann constant = 1.38×10-23 kg m2 s-2 K-1


T = temperature required to trigger the reaction


∴ 



Hence, the gas must be heat up to 1.39×109 K to start fusion.



Question 7.

Obtain the maximum kinetic energy of β-particles, and the radiation frequencies of γ decay in the decay scheme shown in Fig. 13.6. You are given that

m(198Au) = 197.968233 u

m(198Hg) = 197.966760 u



Answer:

From the given diagram γ1 decays from the 1.088 MeV energy level to 0 MeV level.


We have,


E = hν


Where,


h = plank’s constant = 6.6×10-34 J s


ν = frequency


Thus, frequency of radiation radiated by γ1 decay is given by,



(∵ 1 eV = 1.6×10-19 C )


Thus, frequency of radiation radiated by γ2 decay is given by,



(∵ 1 eV = 1.6×10-19 C )


Thus, frequency of radiation radiated by γ3 decay is given by,



(∵ 1 eV = 1.6×10-19 C )


Given,


Mass of , m1 = 2.014102 u


Mass of , m2 = 3.016049 u


The energy of the highest level is given by,



Where,


Δm = Mass defect (or) mass lost during reaction


c = speed of light


∴ E = (197.968233 – 197.966760) u×c2


= 0.001473 u×931.5 MeV/c2


= 1.3720995 MeV


Since, β1 decays from maximum level to 1.088 MeV level then,


Kinetic energy of the β1 particle = (1.3720995 – 1.088) MeV


= 0.2840995 MeV


Since, β2 decays from maximum level to 0.412 MeV level then,


Kinetic energy of the β2 particle = (1.3720995 – 0.412) MeV


= 0.9600995 MeV



Question 8.

Calculate and compare the energy released by a) fusion of 1.0 kg of hydrogen deep within Sun and b) the fission of 1.0 kg of 235U in a fission reactor.


Answer:

a) Mass of hydrogen, m = 1kg = 1000 g


Since,1 mole of hydrogen contains 6.023×1023 atoms which is equivalent to 1 g of hydrogen then, 1kg of hydrogen contains,


N = 6.023×1023×1000 = 6.023×1026 atoms


In sun, 4 hydrogen atoms,  combine to form one helium atom,  in fusion process which releases 26 MeV of energy.


Thus,


Energy released from fusion of 1kg of hydrogen is,


 ……………….(1)



b) Mass of uranium, m = 1kg = 1000 g


Since,1 mole of Uranium contains 6.023×1023 atoms which is equivalent to 235 g of Uranium then, 1kg of Uranium contains,


N = 


During fission reaction of 1 atom of  releases 200 MeV of energy.


Thus,


Energy released from fission of 1kg of Uranium is,


 ………..(2)


Divide (1) by (2) we get,



Hence, Fusion reaction occurred in Sun releases 8 times more energy than energy released during the fission reaction of Uranium.



Question 9.

Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e. conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of 23592U to be about 200MeV.


Answer:

By 2020 India need 2,00,000 MW of electric energy per year, out of which 10% of energy is to be obtained from nuclear power plants.


Thus, Nuclear power plants should produce,



= 20,000×106×365×24×60×60 J


= 6.3072×1017 J of energy


Given that Fission of one  releases 200 MeV of energy.


200 MeV = 200×106×1.6×10-19 J = 3.2×10-11 J


Since, reactor is 25% efficient, total energy produced per atom,


Ea = 0.25×3.2×10-11 J = 8×10-12 J


∴ Number of atoms to be fission to produce required energy,



We know that, 235 g Uranium contains 6.023×1023 atoms.


Thus, for 7.884×1028 atoms we need,



Hence, India need 3.076×104 Kg of uranium in 2020 to produce sufficient electricity.


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