Nuclei Class 12th Physics Part Ii CBSE Solution

Class 12^th 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…

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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,

M_Li = 

=  = 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 = m_p + m_n-m_N

And Binding energy is given by E_B = 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 = m_p + m_n-m_Fe

And Binding energy is given by E_B = 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 = m_p + m_n-m_Bi

And Binding energy is given by E_B = 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,

N_A = 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 = m_p + m_n-m_Fe

And Binding energy is given by E_B = 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/T_1/2) 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, RA^1/3 (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 m_e 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 × m_e)] × c²

We know, m_e = 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 = [ m_A + m_b– m_C – m_d]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 2^nd 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 = R₀A^1/3, 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 = R₀A^1/3 (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.

Q_A = 

= 

And similarly,

Q_B = m

= 

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

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

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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, m_avg = 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 (T_1/2 = 14.3d) and ³³₁₅P (T_1/2 = 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 – 92m_e (∵ atomic number 92)

 = m₂ – 58m_e (∵ atomic number 58)

 = m₃ – 44m_e (∵ 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,

E_a = 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.