##### Class 12^{th} Physics Part Ii CBSE Solution

**Exercises**- Two stable isotopes of lithium^6 3Li and^7 3Li have respective abundances of 7.5% and…
- Boron has two stable isotopes,^10 5B and^11 5B. Their respective masses are 10.01294 u and…
- The three stable isotopes of neon: 2010Ne,^21 10Ne and^22 10Ne, have respective abundances…
- Obtain the binding energy (in MeV) of a nitrogen nucleus (^14 7N) given m (^14 7N) =…
- Obtain the binding energy of the nuclei^56 26Fe and^209 83BI in units of MeV from the…
- A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to…
- Write nuclear reaction equations for (i) α-decay of^226 88Ra (ii) α-decay of^242 94Pu…
- A radioactive isotope has a half-life of T years. How long will it take the activity to…
- The normal activity of living carbon-containing matter is found to be about 15 decays per…
- Obtain the amount of^60 27Co necessary to provide a radioactive source of 8.0 mCi…
- The half-life of^90 38Sr is 28 years. What is the disintegration rate of 15 mg of this…
- Obtain approximately the ratio of the nuclear radii of the gold isotope^197 79Au and the…
- Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of…
- The radionuclide^11 C decays according to The maximum energy of the emitted positron is…
- The nucleus^23 10Ne decays by β- emission. Write down the β-decay equation and determine…
- The Q value of a nuclear reaction A + b → C + d is defined by Q = [mA + mb- mC - md]c^2…
- Suppose, we think of fission of a^56 26Fe nucleus into two equal fragments,^28 13Al. Is…
- The fission properties of^239 94Pu are very similar to those of^235 92U. The average…
- A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much^235 92U did it…
- How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium?…
- Calculate the height of the potential barrier for a head-on collision of two deuterons.…
- From the relation R = R0A1/3, where R0 is a constant and A is the mass number of a…
- For the β + (positron) emission from a nucleus, there is another competing process known…

**Additional Exercises**- In a periodic table, the average atomic mass of magnesium is given as 24.312 u. The…
- The neutron separation energy is defined as the energy required to remove a neutron from…
- A source contains two phosphorous radio nuclides^32 15P (T1/2 = 14.3d) and^33 15P (T1/2 =…
- Under certain circumstances, a nucleus can decay by emitting a particle more massive than…
- Consider the fission by fast neutrons. In one fission event, no neutrons are emitted and…
- Consider the D-T reaction (deuterium-tritium fusion) (a) Calculate the energy released in…
- Obtain the maximum kinetic energy of β-particles, and the radiation frequencies of γ decay…
- Calculate and compare the energy released by a) fusion of 1.0 kg of hydrogen deep within…
- Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten…

**Exercises**

- Two stable isotopes of lithium^6 3Li and^7 3Li have respective abundances of 7.5% and…
- Boron has two stable isotopes,^10 5B and^11 5B. Their respective masses are 10.01294 u and…
- The three stable isotopes of neon: 2010Ne,^21 10Ne and^22 10Ne, have respective abundances…
- Obtain the binding energy (in MeV) of a nitrogen nucleus (^14 7N) given m (^14 7N) =…
- Obtain the binding energy of the nuclei^56 26Fe and^209 83BI in units of MeV from the…
- A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to…
- Write nuclear reaction equations for (i) α-decay of^226 88Ra (ii) α-decay of^242 94Pu…
- A radioactive isotope has a half-life of T years. How long will it take the activity to…
- The normal activity of living carbon-containing matter is found to be about 15 decays per…
- Obtain the amount of^60 27Co necessary to provide a radioactive source of 8.0 mCi…
- The half-life of^90 38Sr is 28 years. What is the disintegration rate of 15 mg of this…
- Obtain approximately the ratio of the nuclear radii of the gold isotope^197 79Au and the…
- Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of…
- The radionuclide^11 C decays according to The maximum energy of the emitted positron is…
- The nucleus^23 10Ne decays by β- emission. Write down the β-decay equation and determine…
- The Q value of a nuclear reaction A + b → C + d is defined by Q = [mA + mb- mC - md]c^2…
- Suppose, we think of fission of a^56 26Fe nucleus into two equal fragments,^28 13Al. Is…
- The fission properties of^239 94Pu are very similar to those of^235 92U. The average…
- A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much^235 92U did it…
- How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium?…
- Calculate the height of the potential barrier for a head-on collision of two deuterons.…
- From the relation R = R0A1/3, where R0 is a constant and A is the mass number of a…
- For the β + (positron) emission from a nucleus, there is another competing process known…

**Additional Exercises**

- In a periodic table, the average atomic mass of magnesium is given as 24.312 u. The…
- The neutron separation energy is defined as the energy required to remove a neutron from…
- A source contains two phosphorous radio nuclides^32 15P (T1/2 = 14.3d) and^33 15P (T1/2 =…
- Under certain circumstances, a nucleus can decay by emitting a particle more massive than…
- Consider the fission by fast neutrons. In one fission event, no neutrons are emitted and…
- Consider the D-T reaction (deuterium-tritium fusion) (a) Calculate the energy released in…
- Obtain the maximum kinetic energy of β-particles, and the radiation frequencies of γ decay…
- Calculate and compare the energy released by a) fusion of 1.0 kg of hydrogen deep within…
- 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 ^{6}_{3}Li and ^{7}_{3}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.

**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, ^{10}_{5}B and ^{11}_{5}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 ^{10}_{5}B and ^{11}_{5}B.

**Answer:**The atomic mass of the two isotopes of Boron (^{10}_{5}B and ^{11}_{5}B) 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 ^{10}B_{5} is 19.89% and that of ^{11}B_{5} is 80.11%

**Question 3.**The three stable isotopes of neon:

^{20}_{10}Ne, ^{21}_{10}Ne and ^{22}_{10}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:**The atomic mass and abundance of ^{20}Ne_{10} is m_{1} = 19.99 u, n_{1} = 90.51%

The atomic mass and abundance of ^{21}Ne_{10} is m_{2} = 20.99 u, n_{2} = 0.27%

The atomic mass and abundance of ^{22}Ne_{10} is m_{3} = 21.99 u, n_{3} = 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 (^{14}_{7}N) given m (^{14}_{7}N) = 14.00307 u

**Answer:**The number of neutron and Proton both are 7 in a ^{14}N_{7} atom.

The mass defect Δm = m_{p} + m_{n}-m_{N}

And Binding energy is given by E_{B} = Δmc^{2}

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/c^{2}

So, E = Δmc^{2} = 0.11236 × 931.5 (MeV/c^{2} × c^{2} ) = 104.66334 MeV

Hence, Binding energy of ^{14}N_{7} = 104.66334 MeV

**Question 5.**Obtain the binding energy of the nuclei ^{56}_{26}Fe and ^{209}_{83}BI in units of MeV from the following data:

m (^{56}_{26}Fe) = 55.934939 u m (^{209}_{83}BI) = 208.980388 u

**Answer:**The number of neutron and Proton both in a ^{56}Fe_{26} atom is respectively 30 and 26.

The mass defect Δm = m_{p} + m_{n}-m_{Fe}

And Binding energy is given by E_{B} = Δmc^{2}

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/c^{2}

So, E = Δmc^{2} = 0.528461 × 931.5(MeV/c^{2}) × c^{2} = 492.26MeV

So, Binding Energy per Nucleon = = = 8.79MeV

The number of neutron and Proton both in a ^{209}Bi_{83} atom is respectively 126 and 83.

The mass defect Δm = m_{p} + m_{n}-m_{Bi}

And Binding energy is given by E_{B} = Δmc^{2}

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/c^{2}

So, E = Δmc^{2} = 1.760877 × 931.5 (Mev/c^{2}) × c^{2} = 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 ^{63}_{29}Cu 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 × 10^{23} atoms /g

Hence, N = = 2.868 × 10^{22}

The number of neutron and Proton both in a ^{63}Cu_{29} atom is respectively 34 and 29.

The mass defect Δm = m_{p} + m_{n}-m_{Fe}

And Binding energy is given by E_{B} = Δmc^{2}

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 × 10^{22} u = 1.69766958 × 10^{22} × 931.5 MeV/c^{2}

Binding Energy = 1.581 × 10^{25}MeV = 1.581 × 10^{25} × 1.6 × 10^{-13} J = 2.5296 × 10^{12} J

Hence, 2.5296 × 10^{12} Joules of energy is required to separate the nucleons of the coin.

**Question 7.**Write nuclear reaction equations for

(i) α-decay of ^{226}_{88}Ra

(ii) α-decay of ^{242}_{94}Pu

(iii) β^{–}-decay of ^{32}_{15}P

(iv) β^{–}-decay of ^{210}_{83}BI

(v) β^{+} -decay of ^{11}_{6}C

vi) β^{+} -decay of ^{43}_{97}Tc

(vii) Electron capture of ^{120}_{54}Xe

**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 N_{0}

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 ^{14}_{6}C present with the stable carbon isotope ^{12}_{6}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) of ^{14}_{6}C, and the measured activity, the age of the specimen can be approximately estimated. This is the principle of ^{14}_{6}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 civilization.

**Answer:**Let N be a number of radioactive carbon found in normal carbon and N_{0} 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 = D_{0} = 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 ^{60}_{27}Co necessary to provide a radioactive source of 8.0 mCi strength. The half-life of ^{60}_{27}Co is 5.3 years.

**Answer:**The Strength a radioactive element is given by =

And, 1Ci = 3.7 × 10^{10}decays/sec;

∵ hence, 8.0 mCi = 8 × 10^{-3} × 3.7 × 10^{10}

= 29.6 × 10^{7} decay/sec (∵ 1 Ci = 3.7 × 10^{10} decay/sec)

Half life of Co^{60} is 5.3 years

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

= 1.67 × 10^{8} s^{-1}

We know,

N = =

Mass of 7.133 × 10^{16} atoms will be = gms

Hence, we need gms of Co^{60}.

**Question 11.**The half-life of ^{90}_{38}Sr 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 × 10^{8} 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 ^{197}_{79}Au and the silver isotope.

**Answer:**__We know that, R____A__^{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) ^{226}_{88}Ra and (b) ^{220}_{86}Rn

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

m (^{222}_{86}Rn) = 220.01137 u, m (^{216}_{84}Po) = 216.00189 u.

**Answer:**a) __Q value of emitted α particle = (Total Initial mass-Total final mass) × c__^{2}

∴ The α particle decay of ^{226}Ra_{88} is given by:

Hence the Q value will be = [226.02540-(222.01750 + 4.002603)] × c^{2}

= 0.005297 u × c^{2}

= 0.005297 × 931.5 Mev

= 4.94MeV

Kinetic energy is given by =

× 4.94 Mev

= 4.85 MeV

b) Similarly, the decay of ^{220}Rn_{86} is given by :

Hence, Q value = [220.01137-(216.00189 + 4.0026)]u × c^{2} = 0.00688u × 931.5 MeV = 6.41MeV

Kinetic energy =

= 6.41 × MeV

= 6.293 MeV

**Question 14.**The radionuclide ^{11}C decays according to

The maximum energy of the emitted positron is 0.960 MeV.

Given the mass values:

m (^{11}_{6}C) = 11.011434 u and m (^{11}_{5}B) = 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 __^{11}Cand 5 m_{e} in the case of ^{11}B 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})] × c^{2}

We know, m_{e} = 0.000548 u

Q = [11.011434-(11.009305 + 2 × 0.000548)] × c^{2}

= 0.001033 u × c^{2}

= 0.962 MeV

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

**Question 15.**The nucleus ^{23}_{10}Ne decays by β^{–} emission. Write down the β-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:

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

m (^{23}_{11}Ne) = 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 × c^{2}

= 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}]c^{2}

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) ^{1}_{1}H + ^{3}_{1}H → ^{2}_{1}H + ^{2}_{1}H

(ii) ^{12}_{6}C + ^{12}_{6}C → ^{20}_{10}Ne + ^{4}_{2}He

Atomic masses are given to be

m (^{2}_{1}H) = 2.014102 u

m (^{3}_{1}H) = 3.016049 u

m (^{12}_{6}C) = 12.000000 u

m (^{20}_{10}Ne) = 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(^{1}H_{1}) + m(^{3}H_{1})-2 × m(^{2}H_{1})] × c^{2}

= (-0.00433u × c^{2}) =

-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(^{12}C_{6})-m(^{20}Ne_{10})-m(^{4}He_{2})] × c^{2}

= (0.004958u) × c^{2}

= 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 ^{56}_{26}Fe nucleus into two equal fragments, ^{28}_{13}Al. Is the fission energetically possible? Argue by working out Q of the process. Given m (^{56}_{26}Fe) = 55.93494 u and m (^{28}_{13}Al) = 27.98191 u.

**Answer:**The fission reaction can be given as :

The Q value for this reaction will be given as = [m()-2 × m()] × c^{2}

= (55.93494-2 × 27.98191)u × c^{2} MeV

= -0.02888u × c^{2} 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 ^{239}_{94}Pu are very similar to those of ^{235}_{92}U. 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 ^{239}_{94}Pu undergo fission?

**Answer:**The atomic mass of ^{239}Pu_{94} is 239.

Hence there are 6.023 × 10^{23} atoms in 239 gm.

So, there are = 2.52 × 10^{24} atoms

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

Hence, E = 180 × 2.52 × 10^{24} Mev = 4.536 × 10^{26}MeV

if all the atoms in 1 kg of pure ^{239}Pu undergo fission then 4.536 × 10^{26}MeV

**Question 19.**A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much ^{235}_{92}U did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of ^{235}_{92}U 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 × 10^{8} 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 × 10^{10} 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:

^{2}_{1}H + ^{2}_{1}H → ^{3}_{2}He + n + 3.27 MeV

**Answer:**The reaction is given As:

Amount of Deuterium fuel = 2.0 kg

∵ 2 gm of deuterium contains 6.023 × 10^{23} atoms

Hence, 2 kg of deuterium contains 6.023 × 10^{23} × 10^{3} 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 × 10^{26} × 10^{6} × 1.6 × 10^{-19}J

= 1.576 × 10^{14} J.

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

Hence, time of glow by this much energy is =

= 1.576 × 10^{12} sec

= (1.576 × 10^{12})/(60 × 60 × 24)

= 4.9 × 10^{4} 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 Є_{0} = permittivity of free space

And = 9 × 10^{9} m^{2}C^{-2}

J

∴

∴ V = 360KeV

Hence, the height of barrier potential is 360 keV.

**Question 22.**From the relation R = R_{0}A^{1/3}, where R_{0} 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 = R___{0}A^{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.

**Question 1.**

Two stable isotopes of lithium ^{6}_{3}Li and ^{7}_{3}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.

**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, ^{10}_{5}B and ^{11}_{5}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 ^{10}_{5}B and ^{11}_{5}B.

**Answer:**

The atomic mass of the two isotopes of Boron (^{10}_{5}B and ^{11}_{5}B) 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 ^{10}B_{5} is 19.89% and that of ^{11}B_{5} is 80.11%

**Question 3.**

The three stable isotopes of neon:^{20}_{10}Ne, ^{21}_{10}Ne and ^{22}_{10}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:**

The atomic mass and abundance of ^{20}Ne_{10} is m_{1} = 19.99 u, n_{1} = 90.51%

The atomic mass and abundance of ^{21}Ne_{10} is m_{2} = 20.99 u, n_{2} = 0.27%

The atomic mass and abundance of ^{22}Ne_{10} is m_{3} = 21.99 u, n_{3} = 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 (^{14}_{7}N) given m (^{14}_{7}N) = 14.00307 u

**Answer:**

The number of neutron and Proton both are 7 in a ^{14}N_{7} atom.

The mass defect Δm = m_{p} + m_{n}-m_{N}

And Binding energy is given by E_{B} = Δmc^{2}

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/c^{2}

So, E = Δmc^{2} = 0.11236 × 931.5 (MeV/c^{2} × c^{2} ) = 104.66334 MeV

Hence, Binding energy of ^{14}N_{7} = 104.66334 MeV

**Question 5.**

Obtain the binding energy of the nuclei ^{56}_{26}Fe and ^{209}_{83}BI in units of MeV from the following data:

m (^{56}_{26}Fe) = 55.934939 u m (^{209}_{83}BI) = 208.980388 u

**Answer:**

The number of neutron and Proton both in a ^{56}Fe_{26} atom is respectively 30 and 26.

The mass defect Δm = m_{p} + m_{n}-m_{Fe}

And Binding energy is given by E_{B} = Δmc^{2}

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/c^{2}

So, E = Δmc^{2} = 0.528461 × 931.5(MeV/c^{2}) × c^{2} = 492.26MeV

So, Binding Energy per Nucleon = = = 8.79MeV

The number of neutron and Proton both in a ^{209}Bi_{83} atom is respectively 126 and 83.

The mass defect Δm = m_{p} + m_{n}-m_{Bi}

And Binding energy is given by E_{B} = Δmc^{2}

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/c^{2}

So, E = Δmc^{2} = 1.760877 × 931.5 (Mev/c^{2}) × c^{2} = 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 ^{63}_{29}Cu 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 × 10^{23} atoms /g

Hence, N = = 2.868 × 10^{22}

The number of neutron and Proton both in a ^{63}Cu_{29} atom is respectively 34 and 29.

The mass defect Δm = m_{p} + m_{n}-m_{Fe}

And Binding energy is given by E_{B} = Δmc^{2}

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 × 10^{22} u = 1.69766958 × 10^{22} × 931.5 MeV/c^{2}

Binding Energy = 1.581 × 10^{25}MeV = 1.581 × 10^{25} × 1.6 × 10^{-13} J = 2.5296 × 10^{12} J

Hence, 2.5296 × 10^{12} Joules of energy is required to separate the nucleons of the coin.

**Question 7.**

Write nuclear reaction equations for

(i) α-decay of ^{226}_{88}Ra

(ii) α-decay of ^{242}_{94}Pu

(iii) β^{–}-decay of ^{32}_{15}P

(iv) β^{–}-decay of ^{210}_{83}BI

(v) β^{+} -decay of ^{11}_{6}C

vi) β^{+} -decay of ^{43}_{97}Tc

(vii) Electron capture of ^{120}_{54}Xe

**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 N_{0}

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 ^{14}_{6}C present with the stable carbon isotope ^{12}_{6}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) of ^{14}_{6}C, and the measured activity, the age of the specimen can be approximately estimated. This is the principle of ^{14}_{6}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 civilization.

**Answer:**

Let N be a number of radioactive carbon found in normal carbon and N_{0} 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 = D_{0} = 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 ^{60}_{27}Co necessary to provide a radioactive source of 8.0 mCi strength. The half-life of ^{60}_{27}Co is 5.3 years.

**Answer:**

The Strength a radioactive element is given by =

And, 1Ci = 3.7 × 10^{10}decays/sec;

∵ hence, 8.0 mCi = 8 × 10^{-3} × 3.7 × 10^{10}

= 29.6 × 10^{7} decay/sec (∵ 1 Ci = 3.7 × 10^{10} decay/sec)

Half life of Co^{60} is 5.3 years

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

= 1.67 × 10^{8} s^{-1}

We know,

N = =

Mass of 7.133 × 10^{16} atoms will be = gms

Hence, we need gms of Co^{60}.

**Question 11.**

The half-life of ^{90}_{38}Sr 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 × 10^{8} 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 ^{197}_{79}Au and the silver isotope.

**Answer:**

__We know that, R____A ^{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) ^{226}_{88}Ra and (b) ^{220}_{86}Rn

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

m (^{222}_{86}Rn) = 220.01137 u, m (^{216}_{84}Po) = 216.00189 u.

**Answer:**

a) __Q value of emitted α particle = (Total Initial mass-Total final mass) × c ^{2}__

∴ The α particle decay of ^{226}Ra_{88} is given by:

Hence the Q value will be = [226.02540-(222.01750 + 4.002603)] × c^{2}

= 0.005297 u × c^{2}

= 0.005297 × 931.5 Mev

= 4.94MeV

Kinetic energy is given by =

× 4.94 Mev

= 4.85 MeV

b) Similarly, the decay of ^{220}Rn_{86} is given by :

Hence, Q value = [220.01137-(216.00189 + 4.0026)]u × c^{2} = 0.00688u × 931.5 MeV = 6.41MeV

Kinetic energy =

= 6.41 × MeV

= 6.293 MeV

**Question 14.**

The radionuclide ^{11}C decays according to

The maximum energy of the emitted positron is 0.960 MeV.

Given the mass values:

m (^{11}_{6}C) = 11.011434 u and m (^{11}_{5}B) = 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 ^{11}Cand 5 m_{e} in the case of ^{11}B 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})] × c^{2}

We know, m_{e} = 0.000548 u

Q = [11.011434-(11.009305 + 2 × 0.000548)] × c^{2}

= 0.001033 u × c^{2}

= 0.962 MeV

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

**Question 15.**

The nucleus ^{23}_{10}Ne decays by β^{–} emission. Write down the β-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:

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

m (^{23}_{11}Ne) = 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 × c^{2}

= 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}]c^{2}

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) ^{1}_{1}H + ^{3}_{1}H → ^{2}_{1}H + ^{2}_{1}H

(ii) ^{12}_{6}C + ^{12}_{6}C → ^{20}_{10}Ne + ^{4}_{2}He

Atomic masses are given to be

m (^{2}_{1}H) = 2.014102 u

m (^{3}_{1}H) = 3.016049 u

m (^{12}_{6}C) = 12.000000 u

m (^{20}_{10}Ne) = 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(^{1}H_{1}) + m(^{3}H_{1})-2 × m(^{2}H_{1})] × c^{2}

= (-0.00433u × c^{2}) =

-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(^{12}C_{6})-m(^{20}Ne_{10})-m(^{4}He_{2})] × c^{2}

= (0.004958u) × c^{2}

= 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 ^{56}_{26}Fe nucleus into two equal fragments, ^{28}_{13}Al. Is the fission energetically possible? Argue by working out Q of the process. Given m (^{56}_{26}Fe) = 55.93494 u and m (^{28}_{13}Al) = 27.98191 u.

**Answer:**

The fission reaction can be given as :

The Q value for this reaction will be given as = [m()-2 × m()] × c^{2}

= (55.93494-2 × 27.98191)u × c^{2} MeV

= -0.02888u × c^{2} 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 ^{239}_{94}Pu are very similar to those of ^{235}_{92}U. 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 ^{239}_{94}Pu undergo fission?

**Answer:**

The atomic mass of ^{239}Pu_{94} is 239.

Hence there are 6.023 × 10^{23} atoms in 239 gm.

So, there are = 2.52 × 10^{24} atoms

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

Hence, E = 180 × 2.52 × 10^{24} Mev = 4.536 × 10^{26}MeV

if all the atoms in 1 kg of pure ^{239}Pu undergo fission then 4.536 × 10^{26}MeV

**Question 19.**

A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much ^{235}_{92}U did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of ^{235}_{92}U 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 × 10^{8} 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 × 10^{10} 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:^{2}_{1}H + ^{2}_{1}H → ^{3}_{2}He + n + 3.27 MeV

**Answer:**

The reaction is given As:

Amount of Deuterium fuel = 2.0 kg

∵ 2 gm of deuterium contains 6.023 × 10^{23} atoms

Hence, 2 kg of deuterium contains 6.023 × 10^{23} × 10^{3} 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 × 10^{26} × 10^{6} × 1.6 × 10^{-19}J

= 1.576 × 10^{14} J.

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

Hence, time of glow by this much energy is =

= 1.576 × 10^{12} sec

= (1.576 × 10^{12})/(60 × 60 × 24)

= 4.9 × 10^{4} 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 Є_{0} = permittivity of free space

And = 9 × 10^{9} m^{2}C^{-2}

J

∴

∴ V = 360KeV

Hence, the height of barrier potential is 360 keV.

**Question 22.**

From the relation R = R_{0}A^{1/3}, where R_{0} 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 = R _{0}A^{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.

###### 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 ^{24}_{12}Mg (23.98504u), ^{25}_{12}Mg (24.98584u) and ^{26}_{12}Mg (25.98259u). The natural abundance of ^{24}_{12}Mg 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 (m_{1}),

Mass of magnesium isotope (m_{2}),

Mass of magnesium isotope (m_{3}),

Abundance of magnesium isotope (n_{1}),

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 (n_{2})

Thus, Abundance of isotope (n_{3}),

We have the relation for the average atomic mass is,

∴

⇒

⇒ -9.272526 u = -0.99675x u

∴ = Abundance of isotope (n_{2}),

Thus, Abundance of isotope (n_{3}),

**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 ^{41}_{20}Ca and ^{27}_{13}Al from the following data:

m(^{40}_{20}Ca) = 39.962591 u

m (^{41}_{20}Ca) = 40.962278 u

m (^{26}_{13}Al) = 25.986895 u

m (^{27}_{13}Al) = 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/c^{2}

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/c^{2}

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 ^{32}_{15}P (T_{1/2} = 14.3d) and ^{33}_{15}P (T_{1/2} = 25.3d). Initially, 10% of the decays come from ^{33}_{15}P. 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 , m_{1} = 223.0185 u

Mass of , m_{2} = 208.98107 u

Mass of , m_{3} = 14.00324 u

∴

But, 1 u = 931.5 MeV/c^{2}

∴

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 , m_{1} = 223.0185 u

Mass of , m_{2} = 219.00948 u

Mass of , m_{3} = 4.00260 u

∴

But, 1 u = 931.5 MeV/c^{2}

∴

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 ^{140}_{58}Ce and ^{99}_{44}Ru. Calculate Q for this fission process. The relevant atomic and particle masses are

m(^{238}_{92}U) = 238.05079 u

m(^{140}_{58}Ce) = 139.90543 u

m(^{99}_{44}Ru) = 98.90594 u

**Answer:**Given that,

Mass of a nucleus , m_{1} = 238.05079 u

Mass of a nucleus , m_{2} = 139.90543 u

Mass of a nucleus , m_{3} = 98.90594 u

Mass of a neutron , m_{4} = 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

= m_{1} – 92m_{e} (∵ atomic number 92)

= m_{2} – 58m_{e} (∵ atomic number 58)

= m_{3} – 44m_{e} (∵ atomic number 44)

= m_{4}

∴

=

= (0.247995)u×c^{2}

But, u = 931.5 MeV/c^{2}

∴ 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 (^{2}_{1}H) = 2.014102 u

m (^{3}_{1}H) = 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 , m_{1} = 2.014102 u

Mass of , m_{2} = 3.016049 u

Mass of , m_{2} = 4.002603 u

Mass of , m_{3} = 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/c^{2}

∴

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

ϵ_{0} = 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 m^{2} s^{-2} K^{-1}

T = temperature required to trigger the reaction

∴

=

Hence, the gas must be heat up to 1.39×10^{9} 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(^{198}Au) = 197.968233 u

m(^{198}Hg) = 197.966760 u

**Answer:**From the given diagram γ_{1} 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 γ_{1} decay is given by,

(∵ 1 eV = 1.6×10^{-19} C )

Thus, frequency of radiation radiated by γ_{2} decay is given by,

(∵ 1 eV = 1.6×10^{-19} C )

Thus, frequency of radiation radiated by γ_{3} decay is given by,

(∵ 1 eV = 1.6×10^{-19} C )

Given,

Mass of , m_{1} = 2.014102 u

Mass of , m_{2} = 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×c^{2}

= 0.001473 u×931.5 MeV/c^{2}

= 1.3720995 MeV

Since, β_{1} decays from maximum level to 1.088 MeV level then,

Kinetic energy of the β_{1} particle = (1.3720995 – 1.088) MeV

= 0.2840995 MeV

Since, β_{2} decays from maximum level to 0.412 MeV level then,

Kinetic energy of the β_{2} 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 ^{235}U in a fission reactor.

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

Since,1 mole of hydrogen contains 6.023×10^{23} atoms which is equivalent to 1 g of hydrogen then, 1kg of hydrogen contains,

N = 6.023×10^{23}×1000 = 6.023×10^{26} 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 6.023×10^{23} 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 ^{235}_{92}U 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,

= 20,000×10^{6}×365×24×60×60 J

= 6.3072×10^{17} J of energy

Given that Fission of one releases 200 MeV of energy.

200 MeV = 200×10^{6}×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 6.023×10^{23} atoms.

Thus, for 7.884×10^{28} atoms we need,

Hence, India need 3.076×10^{4} Kg of uranium in 2020 to produce sufficient electricity.

**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 ^{24}_{12}Mg (23.98504u), ^{25}_{12}Mg (24.98584u) and ^{26}_{12}Mg (25.98259u). The natural abundance of ^{24}_{12}Mg 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 (m_{1}),

Mass of magnesium isotope (m_{2}),

Mass of magnesium isotope (m_{3}),

Abundance of magnesium isotope (n_{1}),

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 (n_{2})

Thus, Abundance of isotope (n_{3}),

We have the relation for the average atomic mass is,

∴

⇒

⇒ -9.272526 u = -0.99675x u

∴ = Abundance of isotope (n_{2}),

Thus, Abundance of isotope (n_{3}),

**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 ^{41}_{20}Ca and ^{27}_{13}Al from the following data:

m(^{40}_{20}Ca) = 39.962591 u

m (^{41}_{20}Ca) = 40.962278 u

m (^{26}_{13}Al) = 25.986895 u

m (^{27}_{13}Al) = 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/c^{2}

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/c^{2}

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 ^{32}_{15}P (T_{1/2} = 14.3d) and ^{33}_{15}P (T_{1/2} = 25.3d). Initially, 10% of the decays come from ^{33}_{15}P. 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 , m_{1} = 223.0185 u

Mass of , m_{2} = 208.98107 u

Mass of , m_{3} = 14.00324 u

∴

But, 1 u = 931.5 MeV/c^{2}

∴

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 , m_{1} = 223.0185 u

Mass of , m_{2} = 219.00948 u

Mass of , m_{3} = 4.00260 u

∴

But, 1 u = 931.5 MeV/c^{2}

∴

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 ^{140}_{58}Ce and ^{99}_{44}Ru. Calculate Q for this fission process. The relevant atomic and particle masses are

m(^{238}_{92}U) = 238.05079 u

m(^{140}_{58}Ce) = 139.90543 u

m(^{99}_{44}Ru) = 98.90594 u

**Answer:**

Given that,

Mass of a nucleus , m_{1} = 238.05079 u

Mass of a nucleus , m_{2} = 139.90543 u

Mass of a nucleus , m_{3} = 98.90594 u

Mass of a neutron , m_{4} = 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

= m_{1} – 92m_{e} (∵ atomic number 92)

= m_{2} – 58m_{e} (∵ atomic number 58)

= m_{3} – 44m_{e} (∵ atomic number 44)

= m_{4}

∴

=

= (0.247995)u×c^{2}

But, u = 931.5 MeV/c^{2}

∴ 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 (^{2}_{1}H) = 2.014102 u

m (^{3}_{1}H) = 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 , m_{1} = 2.014102 u

Mass of , m_{2} = 3.016049 u

Mass of , m_{2} = 4.002603 u

Mass of , m_{3} = 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/c^{2}

∴

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

ϵ_{0} = 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 m^{2} s^{-2} K^{-1}

T = temperature required to trigger the reaction

∴

=

Hence, the gas must be heat up to 1.39×10^{9} 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(^{198}Au) = 197.968233 u

m(^{198}Hg) = 197.966760 u

**Answer:**

From the given diagram γ_{1} 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 γ_{1} decay is given by,

(∵ 1 eV = 1.6×10^{-19} C )

Thus, frequency of radiation radiated by γ_{2} decay is given by,

(∵ 1 eV = 1.6×10^{-19} C )

Thus, frequency of radiation radiated by γ_{3} decay is given by,

(∵ 1 eV = 1.6×10^{-19} C )

Given,

Mass of , m_{1} = 2.014102 u

Mass of , m_{2} = 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×c^{2}

= 0.001473 u×931.5 MeV/c^{2}

= 1.3720995 MeV

Since, β_{1} decays from maximum level to 1.088 MeV level then,

Kinetic energy of the β_{1} particle = (1.3720995 – 1.088) MeV

= 0.2840995 MeV

Since, β_{2} decays from maximum level to 0.412 MeV level then,

Kinetic energy of the β_{2} 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 ^{235}U in a fission reactor.

**Answer:**

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

Since,1 mole of hydrogen contains 6.023×10^{23} atoms which is equivalent to 1 g of hydrogen then, 1kg of hydrogen contains,

N = 6.023×10^{23}×1000 = 6.023×10^{26} 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 6.023×10^{23} 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 ^{235}_{92}U 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,

= 20,000×10^{6}×365×24×60×60 J

= 6.3072×10^{17} J of energy

Given that Fission of one releases 200 MeV of energy.

200 MeV = 200×10^{6}×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 6.023×10^{23} atoms.

Thus, for 7.884×10^{28} atoms we need,

Hence, India need 3.076×10^{4} Kg of uranium in 2020 to produce sufficient electricity.