Physics M.C.Q (1 Marks)

${ }^{12} \mathrm{C}$ 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.