Nuclei

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

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%

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

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

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

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.

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)

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

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.

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.

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

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

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

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.

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

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.

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.

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

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

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.

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.

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.

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.

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),

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,

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,

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.

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.

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.

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

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.

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.