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Option : a. 0.132 y^-1

 \(_{54}^{118}Xe\) → \(_1^0e\) + \(_{53}^{118}I\) 

The age of the wood sample can be determined by radiocarbon dating as ¹⁴C 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, N₀ denotes 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 :

t = \(\frac{2.303}{\lambda}log_{10}\frac{N_o}{N}\) 

Where λ = \(\frac{0.693}{5730y}\) 

= 1.21 x 10^-4 y^-1.

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

Example of mirror nuclei :

\(_1^3H\) and \(_2^3He\)

Option : c. nuclear fusion

\(_Z^AA \overset{\propto-Emission}\longrightarrow\) \(_{Z-2}^{A-4}B\) + \(_2^4He\) 

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.

\(_{90}^{232}Th\) → \(_{82}^{208}Pb\)

The emission of one α-particle decreases the mass number by 4 whereas the emission of βparticles 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 = \(\frac{24}{4}\) = 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.

Nuclear transmutation:

• It is the process of transformation of a stable nucleus into another nucleus which can be stable or unstable. 
• It can occur by the radioactive decay of a nucleus or the reaction of a nucleus with another particle.

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.

e.g,

Step I :

\(_5^{10}B\) + \(_2^4 He\) → \(_7^{13}N\) + \(_0^1n\),

Stable          Radioactive

Step II :

\(_7^{13}N\) → \(_6^{13}C\) + \(_0^1e\)

Radioactive

(Spontaneous emission of position)

Step-I can be considered as nuclear transmutation as it produces a new nuclide \(_7^{13}N\).

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.

Nuclear transmutation is a non-spontaneous (man-made) process.

• An element is considered to be radioactive if the nuclei of its atoms are unstable. 
• That is, when element undergoes nuclear changes (i.e., emission of nuclear particles and gamma radiation), it is said to be radioactive.

The radiations emitted by radioactive elements are as follows:

• Alpha (α) radiations 
• Beta (β) radiations 
• Gamma (γ) radiations

The radius of nucleus is of the order of 10^-15 m whereas that of the outer sphere is of the order of 10^-10 m. The size of outer sphere, is 10⁵ times larger than the nucleus i.e., if we consider the atom of size of football stadium then its nucleus will be the size of a pea.

i. Rate of decay of a radioelement denotes the number of nuclei of its atoms which decay in unit time. It is also called activity of radioelement.

Rate of decay at any time t can be expressed as follows :

Rate of decay (activity) = \(-\frac{dN}{dt}\) 

Where, 

dN is the number of nuclei that decay with in time interval dt.

ii. Minus sign in the expression indicates that the number of nuclei decreases with time. 

Therefore,

dN is a negative quantity. 

But, 

The rate of decay is a positive quantity. 

The negative sign is introduced in the rate expression to make the rate positive.

• Development in earth sciences : Like to understand various geographical changes occurring on earth.
• Development in space technology : To study nuclear reactions in stars which may lead to new discoveries.
• Development in medical sciences : Diagnosis and treatment of various diseases.
• Development in industries : As a potent source of electricity or a power generator.
• Development in agriculture : To study or monitor changes in soil like uptake of nutrients from the soil etc.

When decay is 10 % complete, if N₀ = 100 , then N = 100 – 10 = 90 and t = 20 minutes When decay is 19 % complete, N = 100 – 19 = 81

Substituting these values in formula we get,

For 10% decay completion, \(\lambda\) = \(\frac{2.303}{20}\) log₁₀ \((\frac{100}{90})\).....(i)

For 19% decay completion, \(\lambda\) = \(\frac{2.303}{t}\) log₁₀ \((\frac{100}{81})\).....(ii)

From equation (i) and (ii).

\(\frac{2.303}{20}\) log₁₀ \(\frac{100}{90}\) = .\(\frac{2.303}{t}\) log₁₀ \((\frac{100}{81})\)

 ∴ t log₁₀ \((\frac{100}{90})\) = 20  log₁₀ \((\frac{100}{81})\)

∴ t x 0.0458 = 20 x 0.0915

∴ t = \(\frac{20\times 0.0915}{0.0458}\) = 40 min

• Helium atom is composed of 2 protons and 2 neutrons (or 1 neutron) along with 2 electrons in the outer shell.
• On the other hand, α-particle constitutes 2 protons and 2 neutrons bound together to form a particle which is similar to helium (except presence of electrons).
• Helium is one of the inert gas which is stable (duplet complete) whereas α-particle is unstable and highly reactive.

Chemical reactions:

• Rearrangement of atoms by breaking and forming of chemical bonds. 
• Different isotopes of an element have same behaviour. 
• Only outer shell electrons take part in the chemical reaction. 
• The chemical reaction is accompanied by relatively small amounts of energy. e.g. chemical combustion of 1.0 g methane releases only 56 kJ energy.
• The rates of reaction are influenced by the temperature, pressure, concentration and catalyst.

Nuclear reactions:

• Elements or isotopes of one element are converted into another element in a nuclear reaction. 
• Isotopes of an element behave differently. 
• In addition to electrons, protons, neutrons, other elementary particles may be involved. 
• The nuclear reaction is accompanied by a large amount of energy change, e.g. The nuclear transformation of 1 g of Uranium – 235 release 8.2 × 10⁷ kJ 
• The rate of nuclear reactions is unaffected by temperature, pressure and catalyst.

Chemical reactions :

• Rearrangement of atoms by breaking and forming of chemical bonds.
• Different isotopes of an element have same behaviour.

Nuclear reactions :

• Elements or isotopes of one element are converted into another element in a nuclear reaction.
• Isotopes of an element behave differently.

The order of distance between two protons present in the nucleus is typically of order of 10^-15m.

Given : 

Sun’s mass loss = 4.34 × 10⁹ kg per second

To find : 

Energy radiated per second into space by Sun

Calculation : 

Δm = 4.34 × 10⁹ kg per second

Now, 

1.66 × 10^-27 kg = 1u 

∴ Δm = \(\frac{4.34\times 10^9}{1.66\times 10^{-27}}\) u per second

= 2.614 × 10³⁶u per second

Now, 

1u = 931.4 MeV

2.614 × 10³⁶ u per second = 2.614 × 10³⁶ × 931.4

= 2.435 × 10³⁹ MeV/s

Now,

1 MeV = 1.6022 × 10^-19J and 

1 eV = 1 × 10^-6 MeV

1 MeV = 1.6022 × 10^-13J

= 1.6022 × 10^-16LJ

E = 2.435 × 10 MeV/s × 1.6022 × 10^-16 kJ/MeV

= 3.901 × 10²³ kJ/s

∴ Energy radiated per second into space by Sun is 3.901 × 10²³ kJ/s.

i. The uses of radioactive isotopes in the field of medicine :

a. Polycythaemia : 

The red blood cell count increases in the disease polycythaemia. Phosphorus-32 is used in its treatment.

b. Bone cancer : 

Strontium-89, strontium-90, samarium-153 and radium-223 are used in the treatment of bone cancer.

ii. The uses of radioactive isotopes in the industrial field :

a. Luminescent paint and radioluminescence : 

The radioactive substances radium, promethium, tritium with some phosphorus are used to make certain objects visible in the dark.

e.g. Hands of a clock, krypton-85 is used in HID (High Intensity Discharge) lamps.

b. Use in ceramic articles : 

1. Luminous colours are used to decorate ceramic tiles, utensils, plates, etc. 

2. Uranium oxide was earlier used to colour ceramics.

iii. The uses of radioactive isotopes in the agriculture field :

a. The genes and chromosomes that give seeds its properties like fast growth, higher productivity, etc., can be modified by means of radiation.

b. Onions and potatoes are irradiated with gamma rays from cobalt-60 to prevent their sprouting.

i. The number of stable nuclides with either Z or N odd is about one third of nuclides where both are even.

ii. These nuclides are less stable than those having even number of protons and neutrons. 

iii. In these nuclides one nucleon has no partner and therefore, these nuclides are less stable. 

iv. Further the number of nuclides with odd A are nearly the same, irrespective of Z or N is odd. This indicates that protons and neutrons behave similarly in the respect of stability. 

v. Following table gives the estimate of such nuclides occurring in nature.

For diagnostic purpose, short-lived isotopes are used in order to limit the exposure time to radiation. Note: Diagnostic Radioisotopes are listed below:

• Nuclear potential is the attraction between p-p, n-n and p-n. 
• These attractive forces are independent of the charge on nucleons or attraction between p-p, n-n and p-n are equal. 
• These attractive forces operate over short range within the nucleus. 
• Nuclear potential is responsible for the nuclear stability.

Nuclides which fall outside the belt or stability zone are radioactive nuclides.

 (C) \(^{208}_{82}Pb\)

i. True 

ii. False 

Nuclides with either ‘Z’ or ‘N’ odd are less stable than nuclides having even number of both ‘Z’ and ‘N’.

iii. True

Solar system: It consists of the Sun and planets in which Sun is at the centre of solar system and planets move around it under the force of gravity.

Atomic system: It consists of tiny central core called as nucleus at the centre of atom around which electrons are present. Like in solar system, electrostatic attractions hold subatomic particles in a structure of atom. The nucleus consists of protons and neutrons

Nuclides can be classified into stable and unstable/radioactive nuclides on the basis of nuclear stability.

• Stable nuclides: In this type of nuclides, the number of electrons and the location of nuclei may change in outer sphere but the number of protons and neutrons remain unchanged.
• Radioactive (unstable) nuclides: These nuclides undergo spontaneous change forming new nuclides.

i. False, 

The number of nucleons in C-12 atom is 12. 

ii. True 

iii. True

The term nucleon refers to the sum of protons (p) and neutrons (n) present in atom, e.g. Number of nucleons present in \(^{40}_{20}Ca,\) are 40 (i.e., 20 protons and 20 neutrons). Number of nucleons present in \(^{23}_{11}Na\) are 23 (i.e., 11 protons and 12 neutrons).

• The nuclides with the same number of protons (Z) and neutrons (N) or the same mass number (A) which differ in energy states are called nuclear isomers.
• In this, the isomer of higher energy is said to be in the metastable state which is represented by writing “m” after the mass number.

e.g. Nuclear isomers of cobalt can be represented as, _60mCo and ₆₀Co.

γ-decay: 

i. γ-Radiation is always accompanied with α and β decay processes. 

ii. During γ-radiation, the daughter nucleus is left in energetically excited state which decays to the ground state of product with emission of γ-rays.

For example, \(^{238}_{92}U\) \(\longrightarrow\) \(^{234}_{90}Th\) + \(^{4}_{2}He\) + γ

iii. \(^{238}_{92}U\) emits α-particles of two different energies, 4.147 MeV (23%) and 4.195 MeV (77%).

iv. When α-particles of energy 4.147 MeV are emitted, ²³⁴Th is left in an excited state which de-excites to the ground state with emission of γ-ray photons with energy 0.0048 MeV.

During the formation of nucleus, certain mass is lost. This phenomenon is known as mass defect (Δm).

The exact mass of nucleus is slightly less than sum of the exact masses of the constituent nucleons. This difference is called as mass defect. It is represented by symbol Δm. Formulae: Δm = calculated mass – observed mass

Radiocarbon dating: The technique is used to find the age of historic and archaeological organic samples such as old wood samples and animal or human fossils.

Radioisotope used for carbon dating is ¹⁴C.

i. Radioactive¹⁴C is formed in the upper atmosphere by bombardment of neutrons from cosmic ray on ¹⁴N.

\(^{14}_7N\) + \(^{1}_0n\) \(\longrightarrow\) \(^{14}_6C\) + \(^{1}_1H\)

ii. ¹⁴C combines with atmospheric oxygen to form ¹⁴CO₂ which mixes with ordinary ¹²CO₂ . 

iii. This carbon dioxide is absorbed by plants during photosynthesis.

iv. Animals eat plants which have absorbed a carbon dioxide ( ¹⁴CO₂ + ¹²CO₂). Hence, ¹⁴C becomes a part of plant and animal bodies. v. As long as the plant is alive, the ratio ¹⁴C/ ¹²C remains constant

vi. When the plant dies, photosynthesis will not occur and the ratio ¹⁴C/ ¹²C decreases with the decay of radioactive ¹⁴C which has a halflife 5730 years.

vii. The decay process of ¹⁴C is given below: 

\(^{14}_6C\) \(\longrightarrow\) \(^{14}_7N\) + \(^0_{-1}e\)

viii. The activity (N) of given wood sample and that of fresh sample of live plant (N0) is measured, where, N0 denotes the activity of the given sample at the time of death. 

ix. The age of the given wood sample, can be determined by applying following Formulae:

t = \(\frac{2.303}{\lambda}\) log₁₀ \(\frac{N_0}N\)

Where \(\lambda\) = \(\frac{0.693}{5730\,y}\) = 1.21 x 10^-4 y^-1

Nuclear fusion reaction requires extremely high temperature typically of the order of 108 K.

Nuclear fission:

• It is the process in which a heavy nucleus splits up into two lighter nuclei of nearly equal masses. 
• About 200 MeV of energy is available per fission in case of \(^{235}_{92}U\). 
• The products of nuclear fission are, in general, radioactive.

Nuclear fusion:

• It is the process in which two lighter nuclei combine together to form a heavy nucleus. 
• Energy available per fusion is much less but the energy per unit mass of material is much greater than that for fission of heavy nuclei. 
• The products of fusion are, in general, non-radioactive.

Nuclear fusion will produce relatively more energy per given mass of fuel.

• Nuclear fusion involves the fusion of lighter nuclei to form a heavy nucleus which is accompanied by an enormous amount of energy (heat).
• Fusion reaction requires extremely high temperature typically of the order of 108 K. 

Fission of 1 gram of uranium-235 produces about 24,000 kW/h of energy.

• Nuclear fission of ²³⁵U occurs when nucleus absorbs neutron. When a uranium nucleus absorbs neutron, it breaks into two lighter fragments and releases energy (heat), more neutrons, and other radiation.
• When one uranium 235 nucleus undergoes fission, three neutrons are emitted.
• These neutrons emitted in fission cause more fission of the uranium nuclei which yield more neutrons. These neutrons again bring forth fission producing further neutrons.
• The process continues indefinitely leading to chain reaction which continues even after the removal of bombarding neutrons.

i. It involves α-decay process. 

ii. As uranium undergoes decay by emission of an α-particle (i.e., \(^{4}_{2}He\)), daughter nuclei (in this case thorium) ‘will observe the decrease in atomic number by 2 units and decrease in atomic mass number by 4 units. 

Hence, the given equation is correct.

i. Isotopes: Nuclides which contain same number of protons but different number of neutrons in their nuclei are called as isotopes. 

e.g. \(^{22}_{11}Na\), and \(^{24}_{11}Na\)

ii. Isobars: Nuclides (of different element) which have same mass number but have different number of protons and neutrons in their nuclei are called as isobars. 

OR 

The atoms of different elements having the same mass number but different atomic numbers are called isobars.

e.g.\(^{14}_{6}C\) and \(^{14}_{7}N\)

• The magnitude of electronic charge (e) on the nucleus is +Ze and that of outer sphere is -Ze. Number of protons and number of electrons are always equal in an atom. 
• As a result of this, the charges get nullified, therefore, the atom as a whole is electrically neutral.

Expression for nuclear binding energy: 

i. Consider a nuclide \(^A_ZX\) that contains Z protons and (A – Z) neutrons. Suppose the mass of the nuclide is m. The mass of proton is m_p and that of neutron is mn.

ii. Total mass = (A – Z)m_n + Zm_p + Zm_e …..(1)

Δm = [(A – Z)m_n + Zm_p + Zm_e] – m

= [(A – Z)m_n + Z(m_p + m_e] – m

= [(A – Z)m_n + Zm_H] – m …..(2)

Where (m_p + m_e) = m_H = mass of H atom.

Thus, (Δm) = [Zm_p + (A – Z)m_n] – m

Where Z = atomic number 

A = mass number 

(A – Z) = neutron number

m_p and m_n = masses of proton and neutron, respectively 

m = mass of nuclide

iii. The mass defect, Δm is related to binding energy of nucleus by Einstein’s equation, ΔE = Δm × c²

Where, ΔE = Binding energy, Δm = mass defect. 

iv. Nuclear energy is measured in million electro volt (MeV).

v. The total binding energy is then given by, B.E. = Δm (u) × 931.4 

Where 1.00 u = 931.4 MeV 

B.E. = 931.4 [Zm_H + (A – Z)m_n – m] ……(3)

Total binding energy of nucleus containing A nucleons is the B.E. 

vi. The binding energy per nucleon is then given by, 

\(\bar{B}\) = B.E./A

Correct option is (D) 2p + 2n

Equation for the decay constant is given as,

λ = \(\frac{2.303}{t}\) log₁₀\(\frac{N_0}{N}\) …(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,

λ = \(\frac{2.303}{t_{1/2}}\) log₁₀\(\frac{N_0}{\frac{N_0}{2}}\) 

= \(\frac{2.303}{t_{1/2}}\)log₁₀² 

= \(\frac{2.303}{t_{1/2}}\) x 0.3010 

= \(\frac{0.693}{t_{1/2}}\)

Hence,

λ = \(\frac{0.693}{t_{1/2}}\) or t_1/2 =\(\frac{0.693}{\lambda}\) 

The rate of decay of a radioelement at any instant is proportional to the number of nuclei (atoms) present at that instant. It can be represented as,

\(-\frac{dN}{dt}\) ∝ N or \(-\frac{dN}{dt}\) = λN.....(i)

Where, \(-\frac{dN}{dt}\) = Rate of decay at any time, t 

λ = Decay constant

N = Number of nuclei (atoms) present at time, t 

From equation (i),

λ = \(\frac{(-\frac{dN}{dt})}N\) or λ = \(-\frac{dN}{dt}\) x \(\frac{1}N\)

Decay constant (λ) is the fraction of nuclei decaying in unit time. 

OR 

It is the ratio of the amount of substance disintegrated per unit time to the amount of substance present at that time.

Given :

t_1/2 = 900 minutes

To find : λ

Formula :

λ = \(\frac{0.693}{t_{1/2}}\) 

Calculation :

λ = \(\frac{0.693}{t_{1/2}}\) 

= \(\frac{0.693}{900\,min}\) 

λ = 7.7 x 10^-4 min^-1

∴ Decay constant for ²⁴Na is 7.7 x 10^-4 min^-1.