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JEE Advanced 2023 Revision Notes for Nuclear Chemistry

Last updated date: 19th Mar 2023
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The discovery of the nucleus and other constitutional features of an atom have led to the foundation of a new chapter or domain in physical chemistry, nuclear chemistry. Nuclear chemistry teaches us about the different constituents of the nucleus of an atom and their features. This chapter elaborates on the phenomenon of radioactivity using different scientific principles. To understand the fundamental concepts, use the Nuclear Chemistry IIT JEE notes.

These notes have been compiled by the experts in a simpler version. Learn how the experts have explained the concepts of nuclear chemistry using various fundamental principles and described the laws. It has become a common tool for various disciplines in different domains or subjects. To understand its importance, proceed to use the revision notes.

Competitive Exams after 12th Science

Importance of Physical Chemistry: Nuclear Chemistry

Nuclear chemistry is a subdomain that teaches students about the various phenomena happening inside the nucleus of an atom. There are certain changes occurring inside the nucleus resulting in energy emission and chemical changes.

This subject is also known as radiochemistry. It involves the study of the universe and its components. The development of medicine using radioactive components also involves this part of physical chemistry.

This chapter is important as it teaches the different types of nuclear radiation. This radiation either contains particles or energy. The explanation of all kinds of radioactive rays will be explained in this chapter.

Studying this chapter will also help students to complete the syllabus of Nuclear Chemistry JEE Advanced. They will also get help to complete chapters related to nuclear physics. They will understand the features of fission and fusion reactions.

The importance of this chapter reflects in the explanation of applications of nuclear chemistry in different industrial sectors such as agriculture, radio-labelling, consumer product formulation, smoke detectors, carbon dating, pest control, instrument inspection, etc.

Benefits of Vedantu’s Nuclear Chemistry IIT JEE Notes

• These notes have been compiled by the top subject matter experts of Vedantu. The simplification of all the concepts will enable you to complete learning this chapter faster. The explanation of different radioactivity phenomena will help you to cover the syllabus better.

• Make your study time faster and easier by using the notes. Once you are done preparing this chapter, use these notes to revise and recall the scientific principles. These concise notes will help you recall and prepare the right answers.

• Save your time by using the notes to resolve your doubts and strengthen your conceptual foundation. Find out how the experts have answered the sample questions and progress with your JEE Advanced preparation.

Get the free PDF version of these notes and complete your study material for this chapter. Reduce your preparation time and develop excellent answering skills. Learn how to analyse the fundamental questions coming in the JEE Advanced exams from the explanation given by the experts. Use these notes to recall the concepts and scientific principles faster and define a better exam strategy to score more.

Students can use these Nuclear Chemistry IIT JEE notes at your convenience and make your study sessions more productive. Find out how nuclear chemistry is applied in different industrial sectors. Make your concepts clear with these notes and revise the complete chapter quickly.

JEE Advanced Revision Notes Chemistry Nuclear Chemistry

Nuclear chemistry involves the study of a branch of chemistry that deals with the composition of the atomic nuclei and nuclear forces.

Nuclear Forces:

• Protons and neutrons which live in the center are called nucleons, and forces binding them in the center, are called nuclear forces.

• These are low-lying range forces operating over veritably small distances 1 fermi = $10^{-15}~m$.

• These forces are ${10^{21}}$ times stronger than the electrostatic forces.

• ${n^o}$ and ${p^ + }$ are held together by the veritably rapid-fire exchange of nuclear patches, called mesons.

• Mesons may be positively charged( π), negatively charged, $\left( {{\pi ^ - }} \right)$or neutral( ${\pi ^0}$).

$H_1^1 + \pi _{ - 1}^0 \to n_0^1$

$n_0^1 + \pi _1^0 \to H_1^1$

When a particular nucleus emits the radiation spontaneously, such certain nuclei are known to be radioactive and this phenomenon of disintegration of nuclei spontaneously is called radioactivity.

• Alpha Rays: The rays which bend towards the negative plate by carrying the positive charge.

• Beta Rays: The rays which bend towards the positive plate by carrying the negative charge.

• Gamma Rays: These are uncharged rays that pass straight through the electric field.

Properties of Alpha, Beta, and Gamma Rays:

 Property α - Rays β -Rays γ - Rays Nature Positively charge He nuclei $\left( {H{e^{2 + }}} \right)$ Negatively charge He nuclei $\left( {_{ - 1}{\beta ^0}} \right)$ Electromagnetic radiation Mass 4u 1/1834 u Negligible Velocity 5-10% of the velocity of light. 96-99% of the velocity of light. Same as the speed of the light. Charge +2 unit -1 unit Zero Kinetic energy Very high High Low Penetrating power Small Medium High Ionising power Maximum Moderate Minimum Effect of ZnS Produce luminosity Little effect Very less effect

• The α- particle which is similar to helium nucleus includes two protons and two neutrons.

• The emission part of an alpha particle from an atom’s nucleus is called α-radiation.

When an atom emits an α-particle, the mass of atoms decreases by 4 units.

$U_{92}^{238} \to He_2^4 + Th_{90}^{234}$.

It is the transformation of a neutron into a proton through the emission of an electron. Or the reverse process is transforming a proton into a neutron through the emission of a positron(similar to an electron, but with a positive charge). During this emission, when an atom emits β- particles, the mass of the atom will not change. This resulted in the increase of atomic number by one.

$C_6^{14} \to e_{ - 1}^0 + N_7^{14}$

It involves the emigration of electromagnetic energy from a atoms’s nucleus.

During gamma radiation, no particles are emitted; therefore, it doesn't beget the vacillation of atoms.

$Co_{27}^{60} \to Ni_{28}^{60} + e_{ - 1}^0 + 2\gamma _0^0$

The process of converting one radio-active nucleus into another by emitting the 𝛂, 𝛃, and 𝛄 rays.

• Alpha Decay: The emission of an alpha particle, decreases the atomic number by 2 units and mass number by 4 units.

$U_{92}^{238} \to He_2^4 + Th_{90}^{234}$

Number of $\alpha$-particle emitted= $\dfrac{{{\text{Change in mass number}}}}{{\text{4}}}$

• Beta Decay: Emission of $\beta -$particle, increases the atomic number by 1 unit but does not affect the mass.

This emission gives isotopes.

$C_6^{14} \to e_{ - 1}^0 + N_7^{14}$

A number of $\beta-$ particles emitted =  ${\text{2} \times {Number~of~\alpha - particles - (change~in~atomic ~number)}}$

• Gamma Decay: Emission of γ-rays does not affect the atomic number as well as mass number.

Rate of Disintegration:

The number of atoms of radioactive elements that disintegrate in a unit of time.

Rate of decay = $\dfrac{{ - dN}}{{dt}}\alpha N$

(Or) $\dfrac{{ - dN}}{{dt}} = kN$

Where, k= decay constant

$k = \dfrac{{2.303}}{t}\log \dfrac{{{N_0}}}{{{N_t}}}$

Here, ${N_0}$=number of atoms initially

${N_t}$= a number of atoms after time t.

Half-life Period $\left( {{t_{\dfrac{1}{2}}}} \right)$:

The amount of time required by ab radioactive substance (or one half the atoms) to disintegrate or transform into a different substance.

${t_{\dfrac{1}{2}}} = 0.693/k$

Half life is related to the total time as, $T = n \times {t_{\dfrac{1}{2}}}$

Where, n is calculated from the relation ${N_t} = {N_0}{\left( {\dfrac{1}{2}} \right)^n}$

${N_0}$= initial amount.

${N_1}$= amount after time T.

n = number of half-lives.

Average Life $\left( \lambda \right)$

$\lambda = \dfrac{1}{k}$

$\dfrac{1}{k} = \dfrac{1}{{0.693/{t_{1/2}}}} = 1.44 \times {t_{1/2}}$

Activity is defined as the number of disintegrations occurring in a radioactive substance per second.

Higher is the activity of a substance, faster will be its disintegration.

Activity = kN

$= \dfrac{{{\text{k} \times { wt}}{\text{. of element} \times }{{\text{N}}_{\text{A}}}}}{{{\text{at}}{\text{. wt}}{\text{. of element}}}}$

(${N_A}$=Avogadro’s number =$6.023 \times {10^{23}}$).

 Series 4n 4n+1 4n+2 4n+3 Name Thorium Neptunium Uranium Actinium Parent element $TH_{90}^{232}$ $Pu_{94}^{241}$ $U_{92}^{238}$ $U_{92}^{235}$ Prominent element $Th_{90}^{232}$ $Np_{93}^{237}$ $U_{92}^{238}$ $Ac_{89}^{277}$ End product $Pb_{82}^{208}$ $Bi_{83}^{209}$ $Pb_{82}^{206}$ $Pb_{82}^{207}$ Number of particles lost $\alpha = 6$$\beta = 4 \alpha = 8$$\beta = 5$ $\alpha = 8$$\beta = 6 \alpha = 7$$\beta = 4$

Nuclear Reactions:

• Nuclear Fission: Nucleus in the atoms splits into smaller parts releasing a huge amount of energy in the process.

• Nuclear Fusion: The reaction in which two or more elements fuse together to form a large element, by releasing a large amount of energy in the process.

• Estimation of age (Dating Technique)

Carbon dating technique.

Uranium dating technique.

• Medical use

In therapeutic procedures.

Imaging procedure.

Solved Questions:

1) Two radioactive materials A and B have decay constants 10λ and λ, respectively. If initially, they have the same number of nuclei, then the ratio of the number of nuclei of A to that of B will be 1/e after a time (t). Find the value of t.

Options:

1) $\dfrac{1}{9}\lambda$

2) $\dfrac{{11}}{{10}}\lambda$

3) $\dfrac{1}{{10}}\lambda$

4) $\dfrac{1}{{11}}\lambda$

Ans:

1

Solution:

$N = {N_0}{e^{ - \lambda /t}}$

So, ${{\text{N}}_{\text{1}}}{\text{ = }}{{\text{N}}_{\text{0}}}{{\text{e}}^{{\text{ - 10}\lambda /t}}}{\text{ and }}{{\text{N}}_{\text{2}}}{\text{ = }}{{\text{N}}_{\text{0}}}{{\text{e}}^{{\text{ - }\lambda /t}}}$

$\left( {\dfrac{1}{e}} \right) = \dfrac{{{N_1}}}{{{N_2}}} = \dfrac{{\left( {{N_0}{e^{ - 10\lambda t}}} \right)}}{{{N_0}{e^{ - \lambda t}}}}$

$\left( {\dfrac{1}{e}} \right) = {e^{9\lambda t}} = {e^{ - 1}} = {e^{9 \lambda t}}$

$1 = 9\lambda t$

$t = \dfrac{1}{{9\lambda }}$

2) The half-life period of a radioactive element X is the same as the mean lifetime of another radioactive element Y. Initially, they have the same number of atoms. Then-

Option:

1) X and Y decay at the same rate always

2) X will decay faster than Y

3) Y will decay faster than X

4) X and Y have the same decay rate initially

Ans:

3

Solution:

${T_{\dfrac{1}{2}}}$, half-life of $X = {T_{mean}}$, mean life of Y

Or $\dfrac{{0.069}}{{{\lambda _x}}} = \dfrac{1}{{{\lambda _y}}}$

${\lambda _x} = 0.693{\lambda _y}$

${\lambda _x} < {\lambda _y}$

Rate of decay = ${\lambda_n}$

Initially, number of atoms (N) of both are equal but since ${\lambda _x} < {\lambda _y}$, therefore Y will decay at a faster rate than X.

Hence, Y will decay faster than X.

FAQs on JEE Advanced 2023 Revision Notes for Nuclear Chemistry

1. What is a nucleus in an atom?

A nucleus is the core of an atom made of neutrons and protons. It is the densest part of an atom constituting the positive charges.

2. What is nuclear energy?

The energy emitted from the transformation of one atom to the other is called nuclear energy. This energy is harnessed in nuclear power plants to generate electricity.

3. What is fusion?

The union of two smaller atoms of the same or different elements fuse together to form a complex heavier element is called fusion. During fusion, nuclear energy is emitted.

4. What is fission?

The process of dissociation of a heavier nucleus into simpler ones is called fission. During the fission process, it emits energy while breaking down into at least two simpler nuclei.