Physics M.C.Q (1 Marks)

The nuclear reaction: ${ }_{15} \mathrm{P}^{30} \rightarrow{ }_4 \mathrm{Si}^{30}+{ }_{+1} \mathrm{e}^0$
Thus a positron is emitted during the decay of ${ }_{15} \mathrm{P}^{30}$ into ${ }_{14} \mathrm{Si}^{30}$

The maximum binding energy per nucleon occurs at around mass number $A = 50$, and corresponds to the most stable nuclei. Iron nucleus $F^{56}$ is located close to the peak with a binding energy per nucleon value of approximately $8.8\ MeV$ . It’s one of the most stable nuclides that exist.

Fusion reactions takes place at temperature about $10^7\ K$ it requires this high temperature so that nucleus are moving at very high speed, so that they have high kinetic energy and can overcome the repulsion between nuclei and come together.

$^{22}$Ne decays
$\alpha$ particle $= He^{2+}$
Mass No $= 4$
$p = 2, n = 2$
So, New mass no. $= 22 − 8 = 14$
Atomic No. $= 10 − 4 = 6$
So, the new element is ${6^C}^{14}$

Binding energy per nucleon increases with atomic number. The greater the binding energy per nucleon the more stable is the nucleus.
For $_{26} \mathrm{Fe}^{56}$number of nucleons is $56.$
This is most stable nucleus, since maximum energy is needed to pull a nucleon away from it.

Fusion reaction takes place at temperatures around $10^7k$. It requires this high temperature so that nucleus start moving at rapidly speed, which in turn increases their kinetic, so that they overcome the repulsion between them and can come together.

Hydrogen has $3$ isotopes namely. protium $1​\text{H}\\1$, deuterium $2\text{H}\\1$ and tritium $3\text{H}\\1$

$^{57}Co$ is undergoing beta decay, i.e. electron is being produced. But an electron has very less mass $(9.11 \times 10^{-31}kg)$ as compared to the $Co$ atom. Therefore, after $570$ days, even though the atoms undergo large beta decay, the weight of the material in the container will be nearly $10g.$

The half$-$life of a radioactive sample $\Big(\text{t}_{\frac{1}2{}}\Big)$ is defined as the time elapsed before half the active nuclei decays.
Let the initial number of the active nuclei present in the sample be $N_0$.
$\frac{\text{N}_{0}}{2}=\text{N}_{\text{0}}\text{e}^{-\lambda\text{t}_{\frac{1}2{}}}$
$\Rightarrow\text{t}_{\frac{1}{2}}=\frac{\text{In}2}{\lambda}$
Average life of the nuclei, $\text{t}_{\text{av}}=\frac{\text{S}}{\text{N}_{0}}=\frac{1}{\lambda}$
Here, $S$ is the sum of all the lives of all the $N$ nuclei that were active at $t = 0$ and $\lambda$ is the decay constant of the sample.

$\beta^-$ emission is due to decay of neutron in the nucleus $n \rightarrow p + e^-$.

From conservation of energy:
Change in energy $=$ Energy of reactants $−$ Energy of products $− Q>0 ($Endothermic$)$
Therefore, minimum energy of reactants $>Q=11.32\ MeV$

In nuclear physics, a unit used for measurement of mass is unified atomic mass unit, which is denoted by $u.$
It is defined such that
$1\text{u}=\frac{1}{12}\times ($Mass of neutral carbon atom in its ground state$)$
Mass of neutral carbon atom in its ground state $= 12 \times 1u = 12u$
Thus, the mass of neutral carbon atom in its ground state is exactly $12u.$

Nuclear fusion is a reaction in which two or more atomic nuclei come close enough to form one or more different atomic nuclei and subatomic particles $($neutrons and/or protons$).$
The difference in mass between the products and reactants is manifested as the release of large amounts of energy.
A hydrogen bomb derives its energy from this type of nuclear reaction.

Binding energy per nucleon $= 5.6MeV$
No. of nucleon $=$ No. of proton $+$ No. of neutron
$= 3 + 4 = 7$
So, for $7$ Nucleon $= 7 \times 5.6 = 39.2\ MeV$

1. Fusion requires temperatures about $100$ million Kelvin $($approximately six times hotter titan the sun’s core$).$
2. At these temperatures, hydrogen is a plasma, not a gas. Plasma is a high$-$energy state of matter in which all the electrons are stripped from atoms and move freely about.
3. The sun achieves these temperatures by its large mass and the force of gravity compressing this mass in the core. We must use energy from microwaves, lasers and ion particles to achieve these temperatures.

1. The sun uses its mass and the force of gravity to squeeze hydrogen atoms together in its core.
2. We must squeeze hydrogen atoms together by using intense magnetic fields, powerful lasers or ion beams.

Radiocarbon is produced in the atmosphere as result of collision between fast neutrons and nitrogen nuclei present in the atmosphere.
Nuclear reaction is given as:
${ }_7 \mathrm{~N}^{14}+{ }_0 \mathrm{n}^1 \rightarrow{ }_6 \mathrm{C}^{14}+{ }_1 \mathrm{H}^1$

In a Bainbridge mass spectrometer positive rays of the same element produce different traces. The traces correspond to isotopes. Isotopes are atoms of same element. They have same atomic number $($nuclear charge$)$ but different mass number $($number of neutrons$)$. Positive rays $($or anode rays or canal rays$)$ contains ions obtained by knocking out electrons from gaseous atoms.

$\left(\mathrm{NH}_4\right)_2 \mathrm{CO}_3$​ is the chemical formula of ammonium carbonate.
$N = 14 \times (2) = 28$
$H = 1 \times (4\times 2) = 8$
$C = 12 \times 1 = 12$
$O = 16 \times 3 = 48$
Molar mass $= 28 + 8 + 12 + 48 = 96\ g/mol$

Magnetic force acts on a charged particle, due to which it deflects from its path. The magnitude of this force is measured as $\Big|\overrightarrow{\text{F}}\Big|=\Big|\text{q}\Big(\overrightarrow{\text{v}}\times\overrightarrow{\text{B}}\Big)\Big|.$
Here, $q$ is the charge on the particle that is moving with speed v in a uniform magnetic field $B$.
Since alpha, beta$-$plus and beta$-$minus are charged particles, they suffer deflection due to the field applied. On the other hand, gamma rays are photons and due to zero charge, they do not suffer any deflection.

After release of helium, there will be decrease in atomic number by $2$ and mass number by $4.$

Cobalt therapy or cobalt $-60$ therapy is the medical use of gamma rays from the radioisotope cobalt $60$ to treat conditions such as cancer.

Alpha particle has highest mass of the given option which is he rest mass of the alpha particle amounts to $6.64424 \times 10^{-27} \mathrm{~kg}$. Mass of neutron is $1.0086654 a.m.u.$
Mass of electrons $=9.10938291 \times 10^{-31}$ kilograms.
Hence mass of alpha particle is greatest.

${ }^{12} \mathrm{C}$ and ${ }^{14} \mathrm{C}$ are the two isotopes of carbon atom that have same atomic number, but different mass numbers. This means that they have same number of protons and electrons, but different number of neutrons. Therefore, ​${ }^{12} \mathrm{C}$ has $6$ protons, $6$ electrons and 6 neutrons, whereas ​${ }^{14} \mathrm{C}$ has $6$ electrons, $6$ protons and $8$ neutrons.

$^4_2He→^4_2He^{2+}+2e^-$
$^4_2​He^{2+}$ is alpha particle. Because it has charge equal to $+2e$ and mass is four times the mass of one proton.

Mass number of a nucleus is defined as the sum of the number of neutron and protons present in the nucleus, i.e. the number of nucleons in the nucleus, whereas atomic number is equal to the number of protons present. Therefore, the atomic number is smaller than the mass number. But in the nucleus $($like that of hydrogen ${ }^1 \mathrm{H}_1)$, only protons are present. Due to this, the mass number is equal to the atomic number.

$2.5$ neutrons on the average are released by the fission of each Uranium$-235$ nucleus that absorbs a low energy neutron.

The first three options are correct from the definition of Binding energy.
$B.E.$ has nothing to do with $K.E.$ of the nucleons in nucleus.

$\mathrm{Mc}^2+$ Bindingenergy $=\left[(\mathrm{A}-\mathrm{Z}) \mathrm{Mn}+\mathrm{ZMp]} \mathrm{c}^2\right.$
Therefore, mass of nucleus is less than total mass of its free nucleons.

The $14\text{C}\\\ 2$ has more number of neutrons than protons, so it is the radio isotope in this pair.

${ }_2 \mathrm{He}^4+{ }_{13} \mathrm{Al}^{27} \rightarrow{ }_x \mathrm{P}^{30}+{ }_0 \mathrm{n}^1$

The binding energy is the energy released when a nucleus is assembled from its constituent nucleons. It is thus a measure of the amount of energy held within the bonds of the atom and corresponds to the energy required to be put in again to pull the nucleons apart. Hence, the energy equivalent of the mass-defect is called the binding$-$energy of the nucleus.
The larger the nucleus, the greater is the internal repulsive forces due to the greater number of protons and less energy must be applied to remove a nucleon from the nucleus, hence the binding energy is lower. The greater the binding energy, the more stable the atom is.
This variation in the binding energy per nucleon $(\frac{\text{BE}}{\text{A}})$ is easily seen when the average $\frac{\text{BE}}{\text{A}}$​ is plotted versus atomic mass number $(A),$ as shown in the figure, as the atomic mass number increases i.e. the number of particles in a nucleus increases, the total binding energy also increases first and then decreases for $A > 56.$

A freshly prepared radioactive source emits radiation of intensity that is $64$ times the permissible level. This means that it is possible to work safely till $6$ half$-$lives $(as 2^6 = 64)$ of the radioactive source. Since the half$-$life of the source is $2h$, the minimum time after which it would be possible to work safely with this source is $12h.$

In alpha particle decay, the unstable nucleus emits an alpha particle reducing its proton number $Z$ by $4$ and neutron number $N$ by $2$ such that the element gets changed.
$\text{ }^{\text{A}}_{\text{Z}}\text{X}\rightarrow\text{ }^{\text{A}-4}_{\text{Z}-2}\text{Y}+\text{ }^4_2\text{He}$
During $\beta^--\text{decay},$ a neutron is converted to a proton​, an electron and an antineutrino, i.e. an active nucleus gets converted to one of its isobars and hence the element gets changed.
$\text{ }^{\text{A}}_{\text{Z}}\text{X}\rightarrow\text{ }^{\text{A}}_{\text{Z}+1}\text{Y}+\text{e}+\bar{\text{v}}$
During $\beta^+-\text{decay},$ a proton in the nucleus is converted to a neutron​, a positron and a neutrino in order to maintain the stability of the nucleus, i.e. an active nucleus gets converted to one of its isobars and hence the element gets changed.
$\text{ }^{\text{A}}_{\text{Z}}\text{X}\rightarrow\text{ }^{\text{A}}_{\text{Z}-1}\text{Y}+\beta^++\text{v}$
When a nucleus is in higher excited state or has excess of energy, it comes to the ground state in order to become stable and release energy in the form of electromagnetic radiation called gamma ray. Hence, the element in gamma decay doesn't change.

$ { }_1 \mathrm{H}^2+{ }_1 \mathrm{H}^2 \rightarrow{ }_2 \mathrm{He}^4 $ is a fusion reaction because here two smaller nuclei fuse together to form a single stable nuclei.

For fission, energy to be realeased
$E = (\mathrm{BE})_{\text {products }}-(\mathrm{BE})_{\text {reactants }}$
If products have to be more stable than the reactant, the $BE$ per nucleon has to be higher for products.
Hence, it releases the energy and reaction continues.

The two alpha particles are released along with energy when $Li−7$ is bombarded with a proton. It was found that the mass of the two alpha particles weighs less that the original product in the reaction. Now, the mass that was converted to energy is called as Mass defect.
The difference between the expected mass and the actual mass of an isotope is called mass defect.
The expected mass is calculated by adding the masses of protons, neutrons and electrons present.

From conservation of charge
$7 + 2 = x + 1$
$9 = x + 1$
$x = 8$

mass number $=$ no.of protons $+$ no.of neutrons
Given that mass number of $P = 32$
Atomic number $($no. of electrons $=$ no. of protons$) = 17$
Number of neutrons $= 32 − 17 = 15$
Hence, $P$ satisfies the requirement.

Fusion reaction takes place at temperature about $10^7K.$
It requires this high temperature so that nucleus are moving rapidly, so that they have high kinetic energy and can come together by overcoming repulsion between them.

Each isotopes in the nucleus of hydrogen has one proton $(Z = 1)$. but protium has no neutron, deutrium has $1$ neutron and tritium has $2$ neutrons.

Nuclear binding energy accounts for a noticeable difference between the actual mass of an atom's nucleus and its expected mass based on the sum of the masses of its non$-$bound components. The release in energy accounts for the stability of the bound atom.
Quantitatively, mass defect $= \triangle M =M_{\text {protons }}+M_{\text {neutrons }}-M_{\text {atom }}$​

A radioactive substance will emit $\alpha$ radiation and alpha radiation is nothing but nucleus of Helium atom so the atmosphere will be rich with He.

Thermonuclear reaction, fusion of two light atomic nuclei into a single heavier nucleus by a collision of the two interacting particles at extremely high temperatures, with the consequent release of a relatively large amount of energy.
The Sun is a main$-$sequence star, and thus generates its energy by nuclear fusion of hydrogen nuclei into helium. In its core, the Sun fuses $620$ million metric tons of hydrogen each second.
as shown in the figure that hydrogen atom are fusses to helium atom.

The nuclear force between a neutron and proton is the result of the exchange of charged mesons $(\pi^+\pi^-)$ between them.

When an atom undergoes $\beta-$decay the atomic number increases by $1.$
When an atom undergoes $\beta-$decay, one of the neutrons breaks into one proton and one electron. The resultant electron is then ejected out of the nucleus and this is called as the $\beta$ particle.
While the resultant proton stays inside the nucleus which results in increase of atomic number by $1$, whereas the atomic mass remains invariant.

Key concept:
Half life $(T_{1/2})$: Radioactivity is a process due to which a radioactive material spontaneously decays. Time interval in which the mass of a radioactive substance or the number of its atom reduces to half of its initial value is called the half life of the substance.
i.e., if $\text{N}=\frac{\text{N}_0}{2}$
Then $\text{t}=\text{T}_\frac{1}{2}$
Hence from $\text{N}=\text{N}_0\text{e}^{-\lambda\text{t}}$
$\frac{\text{N}_0}{2}=\text{N}_0\text{e}^{-\lambda\Big(\text{T}_\frac{1}{2}\Big)}\Rightarrow\ \text{T}_\frac{1}{2}=\frac{\log_\text{e}2}{\lambda}=\frac{0.693}{\lambda}$
In half$-$life $(t = 1yr)$ of the material on the average half the number of atoms will decay.
Therefore, the containers will in general have different number of atoms of the material, but their average will be approx $5000.$

It is observed that mass of a stable nucleus is always less than the total mass of constituent nucleons.This difference of mass is known as mass defect. When a nucleus is formed from the free nucleons mass defect is released in the form of energy by Einstein's mass$-$energy relation.
This energy is used to bind the nucleons to form a nucleus therefore an equivalent amount of energy is required to split the nucleus into its parts, that is called binding energy of nucleus.

$_{92} V^{238}+n \rightarrow _{92}A^{239}$
$_{92}A^{239}\rightarrow _{93}B^{239}+e$
$_{92}A^{239}\rightarrow _{93}C^{239}+e$
Finding the element $C$ from periodic table
${ }_ {94} \mathrm{Pu}^{239} $

Hydrogen is the lightest element in the universe with atomic number $1$ and so, it has the simplest atomic structure.

Isotope nucleus means that those nucleus has same protons number but different neutrons and mass number. Since chlorine has $17$ protons so its isotope also will have $17$ protons.

If an alpha particle is bombarded on a nitrogen $(N-14)$ nucleus, an oxygen $(O-17)$ nucleus and a proton are released.
According to the conservation of mass and charge,
$^4_2\text{He}+\text{ }^{14}_7\text{N}\rightarrow\text{ }^{17}_6\text{O}+\text{ }^1_1\text{p}$
So, the emitted particle is a proton.

One an amu is $\frac{1}{10}$​ of the mass of one carbon$-12$ atom.
It is equal to $\frac{1}{\text{NA}}$
$\frac{1}{6.022\times10^{23}}$
$= 1.66 \times 10^{-24} \mathrm{~g} $

Sir Einstein's mass$-$energy equation states that mass and energy can be converted into each other by the following relation.
$\mathrm{E}=\mathrm{mc}^2,(\mathrm{c}=\text { speed of light })$
This implies that a small amount of mass contains a lot of energy, which can be proved with an example.
Let we have a mass of $1 \mathrm{~g}=10^{-3} \mathrm{~kg}$, therefore energy produced by it will be:
$E=10^{-3} \times\left(3 \times 10^8\right) 2=9 \times 10^{13} \mathrm{~J}$ which is a vast amount energy produced by only one gram $($small mass$)$ of mass.
Whereas a small amount of energy doesn't give a large amount of mass because for that we have to divide the energy by $\mathrm{c}^2$, which gives a small mass.

Radius of a nucleus with mass number $A$ is given as
$\text{R}=\text{R}_{\text{0}}\text{A}^{\frac{1}{3}}$
Here, $\text{R}_0=1.2\text{fm}$
$\therefore$ Volume of the nucleus $=\frac{4\pi\text{R}^3}{3}=\frac{4\pi\text{R}^3\text{A}}{3}$
This depends on $A.$ With an increase in $A, V$ increases proportionally.
Mass of the nucleus $\simeq\text{Am}_{\text{N}}$
Here, $m_N <$ is the mass of a nucleon.
Therefore, mass of the nucleus also increases with the increasing mass number. Binding energy also depends on mass number $($number of nucleons$)$ as it is the difference between the total mass of the constituent nucleons and the nucleus. Therefore, it also varies with the changing mass number.
On the other hand,
$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$
$=\frac{\text{Am}_{\text{N}}}{\frac{4\pi\text{R}3}{3}}=\frac{\text{Am}_{\text{N}}}{\frac{4\pi\text{R}_0^3\text{A}}{3}}=\frac{\text{m}_{\text{N}}}{\frac{4\pi\text{R}_0^3}{3}}=\frac{3\text{m}_{\text{N}}}{4\pi\text{R}_{0}^3}$
This is independent of $A$ and hence does not change as mass number increases.

In fast chain reaction neutron released in previous fission again strikes $^{235}U$, So size of $^{235}U$ block should be greater than it's critical size.

In nuclear physics, nuclear fusion is a nuclear reaction in which two or more atomic nuclei collide at a very high speed and join to form a new type of atomic nucleus. During this process, matter is not conserved because some of the matter of the fusing nuclei is converted to photons $($energy$).$

$32\text{​P}\\ \ 15$ is the radioactive isotope of phosphorous element.
As it has more number of neutrons than number of protons.

For neutron decay, some mass disappears as neutrons convert to a proton, electron and antineutrino.
$Q=\left(m_n-m_p-m{_v}^--m_e\right) c^2=0.782 \mathrm{MeV}$

Atomic mass $(u)$ of Calcium is $40 u.$

A nucleus is made up of two fundamental particles$-$neutrons and protons. If a nucleus has more number of neutrons than what is needed to have stability, then neutrons decay into protons and if there's an excess of protons, then they decay to form neutrons. Since a neutron has larger rest mass than a proton, the $Q-$value of its decay reaction is positive and a free neutron decays to a proton, while an isolated proton cannot decay to a neutron as the $Q-$value of its decay reaction is negative. Hence, it is physically not possible.

${ }_{86} \mathrm{Rn}^{220} \rightarrow{ }_{84} \mathrm{Po}^{216}+z \mathrm{X}^A$
$Z + 84 = 86$ and $220 = 216 + A$
So, $Z = 2$ and $A = 4$
$2​​\alpha^4$
So, it is $\alpha$ particle.

Lithium atom contains $3$ protons and $3$ neutrons in the nucleus and $3$ valence electrons. When two lithium nuclei are brought together, they repel each other. The attractive nuclear forces being short$-$range are insignificant as compared to the electrostatic repulsion. Thus, the nuclei do not combine to form carbon atom because of coulomb repulsion.

The average binding energy per nucleon is just the total binding energy divided by the number of nucleons. If we consider Na atom, its binding energy is $194\ MeV.$
Its binding energy per nucleon is given by $\frac{194\text{MeV​}}{24}$
$= 8.08\ MeV$

Key Concept:

Isotopes of an element must have same atomic number $(Z)$ but different mass number $A.$
Number of protons is equal to the atomic number.
So, isotopes of an element have same number of protons.
Mass number is equal to the sum of number of protons and neutrons i.e. $A = p + n$
As isotopes of an element have different mass number but same number of protons, thus they must have different number of neutrons.

$^{40}_{19}{\text{K}}$ is the radioactive isotope of Potassium element.
As it has more number of neutrons than number of protons.

When an alpha particle is emitted from the nucleus the nucleus losses two protons and two neutrons. This means that the atomic number is decreased by $2.$

In nuclear fusion reactions, the heavy isotopes of Hydrogen take part like deuterium ${ }_1 H^2$ and tritium ${ }_1 H^3$ as they have extra neutron.

The concept of atomic mass was proposed by William Prout. Early atomic mass theory was proposed by the English chemist William Prout in a series of published papers in $1815$ and $1816$. Known as Prout's Law, Prout suggested that the known elements had atomic weights that were whole number multiples of the atomic mass of hydrogen.

Modern atomic weight scale is based on $C−12.$
The standard unit for expressing the mass of atom is amu $($atomic mass unit$).$
It is equal to $\frac{1}{12}$ of the mass of an atom of carbon$-12$ equal to $1.6605 \times 10^{-19} \mathrm{~g}$.
amu is also called as avogram.
Avogram is a unit of mass and weight equal to one gram divided by the Avogadro's number.

Fusion reaction takes place at $10^7$k
So, the correct choice is $ 3 \times 10^6 \mathrm{~K} $

$2A \rightarrow B$
Possible if $B$ is more stable than $A$
$⟹$ Energy of $B$ is less than two atoms of $A$
$\mathrm{E}_2 < 2 \mathrm{E}_1$​

The atomic weight is a characteristic property of an element and can never vary.
The valency of an element can vary. For example, when the ion is in $+2$ and $+3$ oxidation state, its valency is $2$ and $3$ respectively.

Packing fraction is the mass defect per nucleon i.e. elementary particle in the nucleus. The difference between atomic weight and mass number i.e. mass of elementary particle in the nucleus is known as mass defect. Hence,
$\text{p}=\frac{\Delta\text{m}}{\text{A}}$
$\text{p}=\frac{\text{M - A}}{\text{A}}$

One $\text{a.m.u.}$ is defined as mass of $\frac{1}{12}th$ the mass of one carbon$-12$ atom.

Radioactive isotopes have unstable nucleus which emits energy and particles when it converts to stable form. Cobalt$-60$ is an isotope used to treat Cancer like problems.

Chlorine is a chemical element with the symbol $Cl$ and atomic number $17$. The second$-$lightest of the halogens, it appears between fluorine and bromine in the periodic table and its properties are mostly intermediate between them.

The difference between the sum of the masses of the constituent particles and the mass of an atom is called mass defect.
The energy released due to mass defect was given by Einstein
$E = \triangle mc^2$

$(BE)_{Hc} = 28\ MeV, A$ of $He = 4$
$(BE)_{He} = \frac{28}{4}\ MeV$
$= 7\ MeV$

Fusion reaction is the combining of atom to form heavier nuclei. Fusion reaction requires high temperature. God fusion to occur on earth a, we need a temperature of at least $100$ million degree Celsius. It is also an uncontrollable reaction.

From conservation of mass
$24 + 4 = x + 1$
$x = 28 − 1 = 27$

Protons and neutrons are present inside the nucleus and they exert strong attractive nuclear force on each other. These forces are equal in magnitude, irrespective of the charge present on the nucleons.
$\therefore$ $ \mathrm{F}_{\mathrm{pp}} = \mathrm{F}_{\mathrm{pn}} = \mathrm{F}_{\mathrm{nn}} $
Here,$ \mathrm{F}_{\mathrm{pp}} = \mathrm{F}_{\mathrm{pn}} = \mathrm{F}_{\mathrm{nn}} $
denote the magnitudes of the nuclear force by a proton on a proton, by a proton on a neutron and by a neutron on a neutron, respectively.

Density of water is $= 1000\ kg/m^3$
Density of sun $= 1.4\times 1000$
$= 1.4 \times 10^3 \mathrm{~kg} / \mathrm{m}^3$

The average nuclear radius $(R)$ and the mass number of the element $(A)$ has the following relation:
$\text{R}=\text{R}_{0}\text{A}^{\frac{1}{2}}$
$\frac{\text{R}}{\text{R}_{0}}=\text{A}^{\frac{1}{3}}$
In $\Big(\frac{\text{R}}{\text{R}_0}\Big)=\frac{1}{3}$ In $A$
Therefore, the graph of ln$\Big(\frac{\text{R}}{\text{R}_0}\Big)$ versus ln $A$ is a straight line passing through the origin with slope $\frac{1}{3}.$

The electron emitted in beta radiation may originates from neutron and it increases the atomic number $1.$

Nuclear fusion reactions are the nuclear reactions in which two or smaller $($lighter$)$ nuclei combine to form a bigger $($heavier$)$ nucleus giving off a huge amount of energy.

Two protons exert strong attractive nuclear force and repulsive electrostatic force on each other. Nuclear forces are short range forces existing in the range of a few fms. Therefore, at a separation of 10nm, the electromagnetic force is greater than the nuclear force, i.e. $ \mathrm{F}_{\mathrm{e}}>>\mathrm{F}_{\mathrm{n}} $

Atomic number $=$ number of protons $= 16$
Mass number $=$ Number of protons $+$ number of neutrons
So the mass number of sulfur $= 32$

Hydrogen bomb is based on nuclear fusion. A large amount of nuclear energy is released by fusion of two light elements $($elements with low atomic numbers$).$
In a hydrogen bomb, two isotopes of hydrogen, deuterium $(1$ proton, $1$ neutron$)$ and tritium $(1$ proton, $2$ neutron$)$ are fused to form a nucleus of helium and a neutron.
This fusion releases $17.6\ MeV$ of energy. Also, there is no limit on the amount of the fusion that can occur.

$\beta^-$ decay: $_\text{Z}^\text{A}\text{​X}→_{\text{Z}+1}^{\text{A}}\text{​Y} + _{−1}^0\text{​e}$
Thus, the atomic number is increased in $\beta^-$ decay.

The nucleus of Tritium (${ }_1 \mathrm{H }^3$) contains $1$ proton and $2$ neutrons.
A neutron decays as $\text{n}\rightarrow\ \text{p}+\overline{\text{e}}+\overline{\text{v}},$ the nucleus may have $2$ protons and one neutron i.e., tritium will transform into ${ }_2 \mathrm{He}^3$. It means triton energy is less that of ${ }_2 \mathrm{He}^3$ nucleaus. Tirton energy is less than that of ${ }_2 \mathrm{He}^3$ nucleus, which simple mean transformation is not allowed energetically.

In fusion reaction $0.65\%$ of mass is converted into energy.
So, $0.65\%$ of mass is lost during nuclear fusion.

The $Q$ value for a reaction is the amount of energy released by that reaction.
In beta decay, a typical $Q$ is around $1\ MeV.$

$\mathrm{C}^{13}+$ energy $\rightarrow \mathrm{C}^{12}+\mathrm{n}$
Energy required to remove one neutron $=$ Difference in total binding energy
$= 13 \times 7.74 − 12 \times 7.68\ MeV$
$= 8.46\ MeV$

The atomic number $= Z =$ no. of protons $= p$
The mass no. $= A =$ no. of protons $(p) +$ no. of neutrons$(n) = n + p$

The mass of $1$ amu is equivalent to an energy of $931\ MeV.$

The order of magnitude of nuclear radii is $1$ fermi $= 10^{-15}$ meters.

Binding energy of a nucleus is defined as the energy required to break the nucleus into its constituents. It is also measured as the $Q-$value of the breaking of nucleus, i.e. the difference between the rest energies of reactants $($nucleus$)$ and the products $($nucleons$)$ or the difference between the kinetic energies of the products and the reactants.

Nuclear binding energy can be computed from the difference in mass of a nucleus, and the sum of the masses of the number of free neutrons and protons that make up the nucleus. Once this mass difference, called the mass defect or mass deficiency, is known, Einstein's mass$-$energy equivalence formula can be used to compute the binding energy of any nucleus.
Early nuclear physicists used to refer to computing this value as a "packing fraction" calculation.

Negative beta decay is given as
$\text{n}\rightarrow\text{p + e}^-+\bar{\text{v}}$
Neutron decays to produce proton, electron and anti$-$neutrino.

The mass of an atom expressed in amu is known as atomic mass. atomic mass unit defined as a unit of mass used to express atomic and molecular weights, equal to one$-$twelfth of the mass of an atom of carbon$-12.$

An exothermic reaction is that in which heat is released.
An exothermic reaction is that in which heat is absorbed.
In a nuclear reaction, if the mass of the product nucleus and outgoing particle is less than the mass of the target nucleus and bombarding particle, then energy is released $($exothermic reaction$),$ and $Q-$value is $+ve.$