Maharashtra State Board Class 11 (FYJC / HSC) Chemistry Solutions Chapter 13 Nuclear Chemistry and Radioactivity - Digest Answers | Maharashtra Board Solutions

Q: Question 1: Identify the following nuclides as: isotopes, isobars and isotones.

Step-by-step solution provided.

Q: Question 1: Prepare a chart of comparative properties of the above three types of radiations.

Step-by-step solution provided.

Q: Question 1: 24Mg and 27Al, both undergo (α, n) reactions and the products are radioactive. These emit β particles having positive charge (called positrons). Write balanced nuclear reactions in

Step-by-step solution provided.

Maharashtra State Board 11th Chemistry Solutions Chapter 13 Nuclear Chemistry and Radioactivity

1. Choose the correct option.

Question A.
Identify nuclear fusion reaction

Answer:
Among the given options, reactions (i) and (ii) represent nuclear fusion reactions wherein lighter nuclei combine to form a heavy nucleus.

Question B.
The missing particle from the nuclear reaction is

Answer:
(A)

Question C.
decays with half-life of 5.27 years to produce . What is the decay constant for such radioactive disintegration ?
a. 0.132 y^-1
b. 0.138
c. 29.6 y
d. 13.8%
Answer:
a. 0.132 y^-1

Question D.
The radioactive isotope used in the treatment of Leukemia is
a.⁶⁰Co
b.²²⁶Ra
c.³²P
d.¹³¹I
Answer:
c.³²P

Question E.
The process by which nuclei having low masses are united to form nuclei with large masses is
a. chemical reaction
b. nuclear fission
c. nuclear fusion
d. chain reaction
Answer:
c. nuclear fusion

2. Explain

Question A.
On the basis of even-odd of protons and neutrons, what type of nuclides are most stable ?
Answer:

Question B.
Explain in brief, nuclear fission.
Answer:
i. Nuclear fission: It is a process which involves splitting of the heavy nucleus of an atom into two nearly equal fragments accompanied by release of the large amount of energy.
e.g. Nuclear fission of²³⁵U

ii. When a uranium nucleus absorbs neutron, it breaks into two lighter fragments and releases energy (heat), more neutrons, and other radiation. This can be given as,

iii. Characteristics of nuclear fission reactions:

Note:

Question C.
The nuclides with odd number of both protons and neutrons are the least stable. Why ?
Answer:

Question D.
Referring the stabilty belt of stable nuclides, which nuclides are β^–and β⁺emitters ? Why ?
Answer:

Question E.
Explain with an example each nuclear transmutation and artifiacial radioactivity. What is the difference between them ?
Answer:
i. Nuclear transmutation: It involves transformation of a stable nucleus into another nucleus takes place which can be either stable or unstable.
ii. Artificial (induced) radioactivity: It is nuclear transmutation where the product nucleus is radioactive. The product nucleus decays spontaneously with emission of radiation and particles.

Step-I can be considered as nuclear transmutation as it produces a new nuclide .
However, the new nuclide is unstable (radioactive). Hence, step-I involves artificial (induced) radioactivity. Thus, in artificial transmutation, a stable element is collided with high speed particles to form another radioactive element.

Question F.
What is binding energy per nucleon ? Explain with the help of diagram how binding energy per nucleon affects nuclear stability ?
Answer:

i. Binding energy per nucleon (), for nucleus containing (A) nucleons with binding energy (B.E.) is given as,
= B.E./A
ii. Mean binding energy per nucleon () for the most stable isotopes as a function of mass number is shown above. This plot leads to the following inferences:
a. Light nuclides: (A < 30)
The peaks with A values in multiples of 4. For example, are more stable.
b. Medium mass nuclides: (30 < A < 90)
increases typically from 8 MeV for A = 16 to nearly 8.3 MeV for A between 28 and 32 and it remains nearly constant 8.5 MeV beyond this and shows a broad maximum. The nuclides falling on the maximum are most stable which turns possess high values.⁵⁶Fe with value of 8.79 MeV is the most stable.
c. Heavy nuclides (A > 90)
decreases from maximum 8.79 MeV to 7.7 MeV for A ≅ 210, 209Bi is the stable nuclide. Beyond this, all nuclides are radioactive (α-emitters).

Question G.
Explain with example α-decay.
Answer:
i. The emission of α-particle from the nuclei of an radioelement is called α-decay.
ii. The charge on an α-particle is +2 with a mass of 4 u.
It is identical with helium nucleus and hence an α-particle is designated as .
iii. In the α-decay process, the parent nucleus emits an α-particle and produces daughter nucleus Y. The parent nucleus thus loses two protons (charge +2) and two neutrons. The total mass lost is 4 u. The daughter nucleus will therefore, have mass 4 units less and charge 2 units less than its parent.
iv. General equation for α-decay process can be given as:

In α-decay process of radium, radon (daughter nuclei) is formed with loses of two protons (charge +2) and two neutrons. The total mass lost is 4 u.
Thus, radon has a mass of 4 units less and charge 2 units less than its parent radium.

Question H.
Energy produced in nuclear fusion is much larger than that produced in nuclear fission. Why is it difficult to use fusion to produce energy ?
Answer:

Question I.
How does N/Z ratio affect the nuclear stability ? Explain with a suitable diagram.
Answer:

[Note: Atoms with unstable nuclei are radioactive (exhibit radioactivity). To become more stable, the nuclei undergo radioactive decay.]

Question J.
You are given a very old sample of wood. How will you determine its age ?
Answer:
The age of the wood sample can be determined by radiocarbon dating as¹⁴C becomes a part of a plant due to the photosynthesis reaction (i.e., absorption of [¹⁴CO₂+¹²CO₂]).
i. The activity (N) of given wood sample and that of fresh sample of live plant (N₀) is measured, where, N₀denotes the activity of the given sample at the time of death.
ii. The age of the given wood sample. can be determined by applying following Formulae:

Note: The oldest rock found so far in Northern Canada is 3.96 billion years old.

3. Answer the following question

Question A.
Give example of mirror nuclei.
Answer:
Example of mirror nuclei: and

Question B.
Balance the nuclear reaction:

Answer:

Question C.
Name the most stable nuclide known. Write two factors responsible for its stability.
Answer:
The most stable nuclide known is lead ().
Two factors responsible for its stability are as follows:

Question D.
Write relation between decay constant of a radioelement and its half life.
Answer:
Relation between decay constant of a radioelement and its half-life is given as, λ =
Where, λ = Decay constant, t_1/2= Half-life of a radioelement

Question E.
What is the difference between an α-particle and helium atom ?
Answer:

Question F.
Write one point that differentiates nuclear reations from chemical reactions.
Answer:
Chemical reactions:

Nuclear reactions:

Question G.
Write pairs of isotones and one pair of mirror nuclei from the following :

Answer:
Isotones: i.
ii.
Mirror nuclei: Since there are no isobars the given set of nuclides does not contain a pair of mirror nuclei.

Question H.
Derive the relationship between half life and decay constant of a radioelement.
Answer:
Equation for the decay constant is given as,
λ = …(i)
Where, λ = Decay constant
N = Number of nuclei (atoms) present at time t
At t = 0, N = N₀.
Hence, at t = t_1/2, N = N₀/2
Substitution of these values of N and t in equation (i) gives,

Question I.
Represent graphically log₁₀(activity /dps) versus t/s. What is its slope ?
Answer:
Equation for a decay constant (λ) is given as,

Hence, instead if log₁₀N versus t, log₁₀ which is log₁₀(activity) is plotted.
The graph of log₁₀(activity/dps) versus t/s gives a straight line which can be represented as follows:

Question J.
Write two units of radioactivity. How are they interrelated ?
Answer:
The unit of radioactivity is curie (Ci).
1 Ci = 3.7 × 10¹⁰dps
ii. Other unit of radioactivity is Becquerel (Bq).
1 Bq = 1 dps
Thus, 1 Ci = 3.7 × 10¹⁰dps = 3.7 × 10¹⁰Bq

Question K.
Half life of²⁴Na is 900 minutes. What is its decay constant?
Answer:

Question L.
Decay constant of¹⁹⁷Hg is 0.017 h^-1. What is its half life ?
Answer:

Question M.
The total binding energy of⁵⁸Ni is 508 MeV. What is its binding energy per nucleon ?
Answer:
Given: B.E. of⁵⁸Ni = 508 MeV,
A = 58
To find: Binding energy per nucleon

Question N.
Atomic mass of is 31.97 u. If masses of neutron and H atom are 1.0087 u and 1.0078 u respectively. What is the mass defect ?
Answer:
Given: m = 31.97 u, Z = 16, A = 32
m_n= 1.0087 u
m_H= 1.0078 u
To find: Δm
Formula: Δm = Zm_H+ (A – Z)m_n– m
Calculation: Δm = Zm_H+ (A – Z)m_n– m
= 16 × 1.0078 + (16 × 1.0087) – 31.97
= [16.1248 + 16.1392] – 31.97
= 0.294 u
Ans: The mass defect is 0.294 u.

Question O.
Write the fusion reactions occuring in the Sun and stars.
Answer:
Fusion reactions occurring in the Sun and stars are can be represented as,

Question P.
How many α and β – particles are emitted in the trasmutation

Answer:

The emission of one α-particle decreases the mass number by 4 whereas the emission of β-particles has no effect on mass number.
Net decrease in mass number = 232 – 208 = 24.
This decrease is only due to α-particles. Hence, number of α-particles emitted = = 6
Now, the emission of one α-particle decrease the atomic number by 2 and one β-particle emission increases it by 1.
The net decrease in atomic number = 90 – 82 = 8
The emission of 6 α-particles causes decrease in atomic number by 12. However, the actual decrease is only 8. Thus, atomic number increases by 4. This increase is due to emission of 4 β-particles.
Thus, 6 α and 4 β-particles are emitted.

Question Q.
A produces B by α- emission. If B is in the group 16 of periodic table, what is the group of A ?
Answer:

When α-emission occurs, atomic number decreases by 2 and atomic mass number by 4.
Thus, if ‘B’ belongs to group 16 of periodic table, that means outermost orbit will contain 6 electrons.
Thus, ‘A’ will have 8 electrons in its valence shell and it will belong to group 18 of the periodic table.

Question R.
Find the number of α and β- particles emitted in the process

Answer:
The emission of one α-particle decreases the mass number by 4 whereas the emission of β-particles has no effect on mass number.
Net decrease in mass number = 222 – 214 = 8. This decrease is only due to α-particle. Hence, number of α-particle emitted = 8/4 = 2
Now, the emission of one α-particle decreases the atomic number by 2 and one β-particle emission increases it by 1.
The net decrease in atomic number = 86 – 84 = 2
The emission of 2 α-particles causes decrease in atomic number by 4. However, the actual decrease is only 2. It means atomic number increases by 2. This increase is due to emission of 2 β-particles.
Thus, 2 α and 2 β-particles are emitted.

[Note: The above question is modified to include the final decay product so as to determine the number of α-particles and β-particles emitted in the process. Here, the final decay product is assumed to be Po-214.]

4. Solve the problems

Question A.
Half life of¹⁸F is 110 minutes. What fraction of¹⁸F sample decays in 20 minutes ?
Answer:
Given: t_1/2= 110 min
t = 20 min
To find: Fraction of¹⁸F simple that decays

∴ Fraction of¹⁸F sample that decays = 1 – 0.882 = 0.118
Ans: Fraction of¹⁸F sample that decays in 20 minutes is 0.118.

Question B.
Half life of³⁵S is 87.8 d. What percentage of³⁵S sample remains after 180 d ?
Answer:
Given: t_1/2= 87.8 d,
N₀= 100,
t = 180 d
To find: % of³⁵S that remains after 180 days

Question C.
Half life⁶⁷Ga is 78 h. How long will it take to decay 12% of sample of Ga ?
Answer:
Given: t_1/2= 78 h,
N₀= 100,
N = 100 – 12 = 88
To find: t

Question D.
0.5 g Sample of²⁰¹Tl decays to 0.0788 g in 8 days. What is its half life ?
Answer:
Given: N₀= 0.5 g,
N = 0.0788 g,
t = 8 days
To find: t_1/2

Question E.
65% of¹¹¹In sample decays in 4.2 d. What is its half life ?
Answer:
Given: N₀= 100,
N = 100 – 65 = 35,
t = 4.2d
To find: t_1/2

Question F.
Calculate the binding energy per nucleon of whose atomic mass is 83.913 u. (Mass of neutron is 1.0087 u and that of H atom is 1.0078 u).
Answer:
Given: A = 84, Z = 36,
m = 83.913 u
m_n= 1.0087 u
m_H= 1.0078 u
To find: Binding energy per nucleon

Question G.
Calculate the energy in Mev released in the nuclear reaction

Atomic masses : Ir = 173.97 u,
Re = 169.96 u and
He = 4.0026 u
Answer:
Given: m_Ir= 173.97 u
m_Re= 169.96 u
m_He= 4.0026 u
To find: Energy released
Formulae: i. Δm = (mass of¹⁷⁴Ir) – (mass of¹⁷⁰Re + mass of⁴He)
ii. E = Δm × 931.4 MeV
Calculation:i. Δm = (mass of¹⁷⁴Ir) – (mass of¹⁷⁰Re + mass of⁴He)
= 173.97 – (169.96 + 4.0026)
= 7.4 × 10^-3u
ii. E = Δm × 931.4
= 7.4 × 10^-3× 931.4
= 6.89236 MeV ≈ 6.892 MeV
Ans: The energy released in given nuclear reaction is 6.892 MeV.

Question H.
A 3/4 of the original amount of radioisotope decays in 60 minutes. What is its half life ?
Answer:

Question I.
How many – particles are emitted by 0.1 g of²²⁶Ra in one year?
Answer:
Given: t = 1 y,
Amount of sample = 0.1 g
To find: Number of particles emitted

Activity = = λN
= 4.28 × 10^-4× 2.665 × 10²⁰atoms
= 1.141 × 10¹⁷particles/year
Ans: Particles emitted by 0.1 g of²²⁶Ra in one year = 1.141 × 10¹⁷particles/year.
[Note: The half-life of radium is 1620 years. In order to apply appropriate textual concept, we have used this value in calculation.]

Question J.
A sample of³²P initially shows activity of one Curie. After 303 days the activity falls to 1.5× 10⁴dps. What is the half life of³²P ?
Answer:

Question K.
Half life of radon is 3.82 d. By what time would 99.9 % of radon will be decayed.
Answer:
Given: t_1/2= 3.82 d,
N₀= 100
N = 100 – 99.9 = 0.1
To find: t

Question L.
It has been found that the Sun’s mass loss is 4.34 × 10⁹kg per second. How much energy per second would be radiated into space by the Sun ?
Answer:
Given: Sun’s mass loss = 4.34 × 10⁹kg per second
To find: Energy radiated per second into space by Sun
calculation: Δm = 4.34 × 10⁹kg per second
Now, 1.66 × 10^-27kg = 1u
∴ Δm = u per second
= 2.614 × 10³⁶u per second
Now, 1 u = 931.4 MeV
2.614 × 10³⁶u per second = 2.614 × 10³⁶× 931.4
= 2.435 × 10³⁹MeV/s
Now, 1 MeV = 1.6022 × 10^-19J and 1 eV = 1 × 10^-6MeV
1 MeV = 1.6022 × 10^-13J
= 1.6022 × 10^-16LJ
E = 2.435 × 10³⁹MeV/s × 1.6022 × 10^-16kJ/MeV
= 3.901 × 10²³kJ/s
Ans: Energy radiated per second into space by Sun is 3.901 × 10²³kJ/s.

Question M.
A sample of old wood shows 7.0 dps/g. If the fresh sample of tree shows 16.0 dps/g, How old is the given sample of wood ? Half life of¹⁴C 5730 y.
Answer:

Activity :

1. Discuss five applications of radioactivity for peaceful purpose.
Answer:

[Note: Students can use above points are reference to discuss topic in class].

2. Organize a trip to Bhabha Atomic Reasearch Centre, Mumbai to learn about nuclear reactor. This will have to be organized through your college.
Answer:
Students are expected to visit the place to understand more about nuclear reactors.

11th Chemistry Digest Chapter 13 Nuclear Chemistry and Radioactivity Intext Questions and Answers

Do you know? (Textbook Page no. 190)

Question 1 Maharashtra Board Solution

How small is the nucleus in comparison to the rest of the atom?