Nucleus (Stability and Reaction) Questions in English

(D) The rate of decay of radioactive species is a nuclear process and is independent of all external factors such as $Temperature$,$Pressure$,or $Chemical \text{ } environment$. Therefore,the activity remains unchanged by these factors.

(D) When ${}^{235}U$ is bombarded with a neutron,it undergoes nuclear fission to produce lighter nuclei such as Barium and Krypton.
The balanced nuclear equation is:
${}_{92}^{235}U + {}_{0}^{1}n \to {}_{56}^{141}Ba + {}_{36}^{92}Kr + 3{}_{0}^{1}n$
Thus,the correct option is $(D)$.

(D) Radioactivity is the process by which an unstable atomic nucleus loses energy by radiation.
It occurs due to the spontaneous decay of unstable nuclei to achieve a more stable configuration.
Therefore,the correct option is $D$.

(D) In nuclear reactors,fissile materials such as $U^{235}$ or $Pu^{239}$ (Plutonium) are used as fuel to sustain the chain reaction.
Therefore,the correct option is $(D)$.

(B) The hydrogen bomb is based on the principle of nuclear fusion. In this process,lighter nuclei (isotopes of hydrogen) combine to form a heavier nucleus,releasing a large amount of energy. The reaction is as follows:
$_1H^2 + _1H^3 \to _2He^4 + _0n^1 + \text{energy}$

(A) In nuclear reactors,moderators are used to slow down the speed of fast-moving neutrons to thermal energies.
Heavy water $(D_2O)$ is commonly used as a moderator because it is effective at slowing down neutrons without absorbing them significantly.
Therefore,the correct option is $(A)$.

(D) The energy produced in a nuclear reaction is given by Einstein's mass-energy equivalence principle,which is expressed by the equation $E = mc^2$,where $E$ is the energy,$m$ is the mass defect,and $c$ is the speed of light in a vacuum. Therefore,the correct option is $(D)$.

(A) In a nuclear fusion reaction,two light nuclei combine to form a heavier,more stable nucleus.
According to the binding energy curve,the binding energy per nucleon increases as the mass number increases for light nuclei (up to $A \approx 56$).
Therefore,the average binding energy per nucleon increases,resulting in the release of energy.

(A) The correct answer is $(A)$.
In a nuclear reactor,control rods are used to regulate the rate of the fission reaction by absorbing excess neutrons.
These rods are typically made of materials with high neutron-absorption cross-sections,such as cadmium $(Cd)$ or boron $(B)$.
Therefore,long rods of $Cd$ are used to absorb the potentially destructive neutrons.

(C) The radioactive isotope $_{6}C^{14}$ is produced in the upper atmosphere by the action of cosmic ray neutrons on $_{7}N^{14}$ atoms.
This isotope is then incorporated into living organisms,and its decay is used for radiocarbon dating.

(C) The fuel used in an atomic pile (nuclear reactor) is typically a fissile material. $Uranium$ $(^{235}U)$ or $Plutonium$ $(^{239}Pu)$ are commonly used as atomic fuels because they undergo nuclear fission to release energy.

(B) The atom bomb is based on the principle of nuclear fission. In this process,a heavy nucleus like $U^{235}$ is bombarded with neutrons,causing it to split into smaller nuclei and releasing a tremendous amount of energy.

(D) Hahn and Strassmann discovered the phenomenon of nuclear fission in $1939$ when they bombarded uranium with neutrons.

(C) The rate of radioactive disintegration is a first-order nuclear process that depends solely on the number of radioactive nuclei present at a given time $(N)$.
It is independent of external physical conditions such as temperature,pressure,or the presence of a vacuum.
Therefore,the rate of disintegration remains unchanged.

(A) The packing fraction is defined as the difference between the isotopic mass $(M)$ and the mass number $(A)$,divided by the mass number $(A)$,multiplied by $10^4$.
Formula: $\text{Packing fraction} = \frac{M - A}{A} \times 10^4$
Given: $M = 55.92066$,$A = 56$
Calculation: $\frac{55.92066 - 56}{56} \times 10^4 = \frac{-0.07934}{56} \times 10^4 = -0.00141678 \times 10^4 = -14.1678 \approx -14.167$

(D) An atom bomb explosion is based on the principle of nuclear fission.
In this process,a heavy nucleus splits into lighter nuclei.
The sum of the masses of the product nuclei is slightly less than the mass of the original nucleus.
This mass difference,known as the mass defect $(\Delta m)$,is converted into energy according to Einstein's mass-energy equivalence equation,$E = \Delta m c^2$.

(A) . $^{14}C$ is a natural radioactive isotope of $^{12}C$ that is continuously produced in the upper atmosphere by cosmic rays.

(B) In a nuclear fission reaction,the total mass of the products is less than the mass of the parent nucleus. This mass difference,known as the mass defect,is converted into energy according to Einstein's mass-energy equivalence principle,$E = \Delta m c^2$.

(A) The large energy released in an atomic bomb explosion is due to nuclear fission,where a heavy nucleus splits into lighter nuclei.
According to Einstein's mass-energy equivalence principle,$E = \Delta m c^2$,the mass of the products is slightly less than the mass of the initial reactants.
This mass defect $(\Delta m)$ is converted into a massive amount of energy.

(B) The given reaction is $_1H^2 + _1H^3 \to _2He^4 + _0n^1 + \text{energy}$.
In this reaction,two light nuclei (deuterium and tritium) combine to form a heavier nucleus (helium) with the release of a large amount of energy.
This process is known as nuclear fusion,which is the principle behind the hydrogen bomb.

(C) India conducted its first underground nuclear test,code-named $Smiling \ Buddha$,on $May \ 18, 1974$,at $Pokhran$ in the state of Rajasthan.
This site was chosen for its remote location and geological suitability for underground testing.

(B) The energy required to separate the nucleons (protons and neutrons) from the nucleus is known as the $Nuclear \ binding \ energy$. This energy is equivalent to the mass defect of the nucleus,which is the difference between the sum of the masses of individual nucleons and the actual mass of the nucleus,as per Einstein's mass-energy equivalence relation,$E = \Delta m c^2$.

(B) The splitting of a heavier atom like that of $^{235}U$ into a number of fragments of much smaller mass by suitable bombardment with sub-atomic particles with the liberation of a huge amount of energy is called nuclear fission.

(C) Radioactivity is a nuclear phenomenon that involves the decay of unstable nuclei.
It is independent of external factors such as temperature,pressure,or chemical combination.
Therefore,the statement that its rate is affected by changes in temperature and/or pressure is incorrect.

(D) . Nuclear fission is a process in which a heavy nucleus splits into two smaller nuclei of comparable mass,accompanied by the release of energy and the emission of elementary nuclear particles such as neutrons.

(A) The huge amount of energy released during atomic fission is due to the conversion of mass into energy,which is described by Einstein's mass-energy equivalence equation,$E = \Delta m c^2$. During the fission process,the sum of the masses of the products is slightly less than the mass of the original nucleus,resulting in a mass defect $(\Delta m)$. This lost mass is released as a large amount of energy.

(A) Mass defect is the measure of binding energy of a nucleus. The binding energy $(BE)$ is calculated using the formula $BE = \Delta m \times c^2$,where $\Delta m$ represents the mass defect.

(B) The correct answer is $(B)$.
In a nuclear reactor,$Cd$ (cadmium) and boron rods are used as control rods.
These rods have the ability to absorb neutrons,thereby regulating the rate of the fission chain reaction.

(A) fusion bomb,also known as a thermonuclear bomb,operates on the principle of nuclear fusion.
In this process,lighter nuclei (typically isotopes of hydrogen like deuterium and tritium) combine at extremely high temperatures and pressures to form a heavier nucleus (helium),releasing a massive amount of energy.

(D) Radioactivity is a nuclear phenomenon that depends on the stability of the nucleus,not on the chemical state of the element.
Since the formation of radium chloride involves only the rearrangement of electrons to form chemical bonds,the nucleus of the radium atom remains unchanged.
Therefore,the radioactivity of the radium atom remains the same regardless of whether it is in its elemental form or combined in a compound.

(C) Nuclear fission is a process in which a heavy nucleus splits into two or more lighter nuclei with the release of energy and neutrons.
Option $C$ represents the fission of a uranium-$235$ nucleus induced by a neutron,which is a classic example of nuclear fission.

(B) The components of a nuclear reactor are matched as follows:
$1$. Moderator: Used to slow down fast neutrons. Graphite is a common moderator.
$2$. Control rods: Used to absorb neutrons and control the rate of fission. Boron is commonly used.
$3$. Fuel rods: Contain the fissile material. Uranium is the standard fuel.
$4$. Coolant: Used to remove heat from the reactor core. Liquid sodium is often used in fast breeder reactors.
Therefore,the correct matching is $1-b, 2-c, 3-a, 4-e$.

(A) The stability of a nucleus depends on its $n/p$ ratio.
Isotopes with an excessive $n/p$ ratio (too many neutrons) are unstable and tend to undergo $\beta^-$ decay to convert a neutron into a proton,thereby decreasing the $n/p$ ratio.
This process involves the emission of an electron $(e^-)$.

(C) The stability of an isotope is often related to its neutron-to-proton ratio ($n/p$ ratio).
For $_{30}Zn^{64}$,the number of protons $(p)$ is $30$ and the number of neutrons $(n)$ is $64 - 30 = 34$.
The $n/p$ ratio is $\frac{34}{30} \approx 1.13$.
For $_{30}Zn^{66}$,$n = 36$,so $n/p = \frac{36}{30} = 1.2$.
For $_{30}Zn^{71}$,$n = 41$,so $n/p = \frac{41}{30} \approx 1.37$.
Since $_{30}Zn^{64}$ has the lowest $n/p$ ratio among the given options and is a naturally occurring stable isotope of Zinc,it is the most stable.

(A) In positron emission,a proton $(p)$ is converted into a neutron $(n)$ and a positron $(e^+)$ is emitted: $p \rightarrow n + e^+$.
Since the total number of nucleons (protons + neutrons) remains constant during this process,the mass number $(A)$ of the resulting isotope remains the same.
Therefore,the correct option is $(A)$.

(C) The nuclear reaction is: $^{35}_{16}S \rightarrow ^{35}_{17}Cl + ^{0}_{-1}e + \bar{\nu}$.
Mass defect $(\Delta m) = \text{mass of } ^{35}S - \text{mass of } ^{35}Cl$.
$\Delta m = 34.96903 \ amu - 34.96885 \ amu = 0.00018 \ amu$.
The energy released $(Q)$ is given by $Q = \Delta m \times 931 \ MeV/amu$.
$Q = 0.00018 \times 931 \ MeV = 0.16758 \ MeV$.

(B) The radius of a nucleus is given by $r = 1.3 \times 10^{-13} \times (A)^{1/3} \ cm$,where $A$ is the mass number.
For $U^{238}$,$r_{U} = 1.3 \times 10^{-13} \times (238)^{1/3} \approx 8.06 \times 10^{-13} \ cm$.
For the $\alpha$-particle $(He^{4})$,$r_{\alpha} = 1.3 \times 10^{-13} \times (4)^{1/3} \approx 2.06 \times 10^{-13} \ cm$.
The total distance $r$ between the centers of the nuclei is $r = r_{U} + r_{\alpha} = 8.06 \times 10^{-13} + 2.06 \times 10^{-13} = 10.12 \times 10^{-13} \ cm$.
The coulombic repulsion energy is $E = \frac{k q_1 q_2}{r}$ (in $CGS$ units,$k=1$).
$E = \frac{(Z_1 e)(Z_2 e)}{r} = \frac{(92 \times 4.8 \times 10^{-10} \ esu) \times (2 \times 4.8 \times 10^{-10} \ esu)}{10.12 \times 10^{-13} \ cm} \ erg$.
$E = \frac{423.936 \times 10^{-20}}{10.12 \times 10^{-13}} \ erg \approx 4.189 \times 10^{-5} \ erg$.
Converting to $eV$ $(1 \ erg = 6.242 \times 10^{11} \ eV)$:
$E = 4.189 \times 10^{-5} \times 6.242 \times 10^{11} \ eV \approx 26.147738 \times 10^4 \ eV$.

(C) The number of atoms in $1 \ g$ of $_{92}U^{235}$ is given by $\frac{1 \ g}{235 \ g/mol} \times 6.023 \times 10^{23} \ \text{atoms/mol} \approx 2.563 \times 10^{21} \ \text{atoms}$.
Energy released per atom is $3.20 \times 10^{-11} \ J$.
Total energy released $= (2.563 \times 10^{21} \ \text{atoms}) \times (3.20 \times 10^{-11} \ J/\text{atom}) = 8.2016 \times 10^{10} \ J$.
Converting to $kJ$: $8.2016 \times 10^{10} \ J = 8.2016 \times 10^7 \ kJ \approx 8.21 \times 10^7 \ kJ$.

(B) The law of conservation of mass states that matter can neither be created nor destroyed in a chemical reaction.
However,in $nuclear$ $reactions$,a small amount of mass is converted into energy according to Einstein's mass-energy equivalence equation,$E = mc^2$.
Therefore,the total mass is not conserved in nuclear reactions,making them an exception to the law of conservation of mass.

(A) For $_{13}^{30}Al$: Number of protons $(Z)$ = $13$ (odd),Number of neutrons $(N)$ = $30 - 13 = 17$ (odd).
For $_{20}^{40}Ca$: Number of protons $(Z)$ = $20$ (even),Number of neutrons $(N)$ = $40 - 20 = 20$ (even).
Nuclides with even numbers of both protons and neutrons are generally more stable than those with odd numbers of both.
Since $_{13}^{30}Al$ has odd-odd nucleons and $_{20}^{40}Ca$ has even-even nucleons,$_{13}^{30}Al$ is less stable.
Statement $2$ is a general rule regarding nuclear stability based on the odd-even nature of nucleons.

(C) Ordinary water absorbs fast-moving neutrons,which leads to the capture of neutrons rather than their moderation,thereby hindering the nuclear fission chain reaction.

(C) The process described is $\beta^-$-decay,where a neutron transforms into a proton and an electron: $_0n^1 \to _{+1}p^1 + _{-1}e^0$.
Since the atomic number $(Z)$ is equal to the number of protons,the transformation of a neutron into a proton increases the atomic number by $1$.
Therefore,the new element formed has a nuclear charge higher by one unit.

(A) The $4n$ series is also known as the Thorium series.
It starts with $_{90}Th^{232}$ and undergoes a series of $\alpha$ and $\beta$ decays.
The stable end product of this radioactive decay series is the lead isotope $_{82}Pb^{208}$.

(C) Radioactive disintegration is a nuclear process that involves changes in the nucleus of an atom,whereas a chemical change involves the rearrangement of electrons and does not affect the nucleus. Therefore,the fundamental difference is that radioactivity is a nuclear process.

(B) The reaction is $_1H^2 + _1H^2 \to _2He^4$.
Total binding energy of reactants = $(2 \times 2 \times x) = 4x$.
Total binding energy of product = $(4 \times 7x) = 28x$.
Energy liberated $(Q)$ = $(\text{Binding energy of product} - \text{Binding energy of reactants}) = 28x - 4x = 24x$.
The ratio of energy liberated to the binding energy per nucleon of $_1H^2$ is $\frac{24x}{x} = 24$.

(C) In a nuclear decay reaction,the sum of mass numbers $(A)$ and the sum of atomic numbers $(Z)$ must be conserved on both sides of the equation.
Given the reaction: ${}_Z^AX \to {}_{6}^{12}C + {}_{+1}^{0}e$.
For mass number $(A)$: $A = 12 + 0 = 12$.
For atomic number $(Z)$: $Z = 6 + 1 = 7$.
Therefore,the values are $A = 12$ and $Z = 7$.

(C) In a nuclear reaction,the total mass number $(A)$ and the total atomic number $(Z)$ must be conserved on both sides of the equation.
Given the reaction: ${}_{Z}^{A}X \to {}_{6}^{12}C + {}_{+1}^{0}e$
Applying conservation of mass number: $A = 12 + 0 = 12$.
Applying conservation of atomic number: $Z = 6 + 1 = 7$.
Therefore,the values are $A = 12$ and $Z = 7$.

(C) The rate of radioactive disintegration is given by the equation: $-\frac{dN}{dt} = \lambda N$.
Here,$-\frac{dN}{dt}$ represents the number of atoms disintegrating per unit time,$\lambda$ is the decay constant,and $N$ is the number of radioactive atoms present.
Radioactive decay is a first-order nuclear process and is independent of external physical conditions like temperature,pressure,or concentration.
Since the decay constant $\lambda$ remains unchanged with temperature,the rate of disintegration depends only on the number of atoms $N$.
If the amount of substance $N$ is increased three times,the rate of disintegration $-\frac{dN}{dt}$ will also increase by a factor of $3$ (i.e.,it becomes triple).

(A) The mass of a nucleus is given by $M = Z m_p + (A-Z) m_n - \text{binding energy}/c^2$.
For $_{10}^{20}Ne$,$Z=10$ and $A=20$,so $M_1 \approx 10 m_p + 10 m_n - \text{binding energy}_1/c^2$.
For $_{20}^{40}Ca$,$Z=20$ and $A=40$,so $M_2 \approx 20 m_p + 20 m_n - \text{binding energy}_2/c^2$.
Since the binding energy per nucleon is higher for $_{20}^{40}Ca$ than for $_{10}^{20}Ne$,the mass defect per nucleon is higher for Calcium.
Therefore,$M_2 < 2 M_1$.
However,in the context of simplified nuclear models where binding energy is neglected,$M_1 = 10(m_p + m_n)$ and $M_2 = 20(m_p + m_n)$,which leads to $M_2 = 2 M_1$.

(D) The binding energy per nucleon is a measure of the stability of a nucleus. The actual order of binding energy per nucleon for these isotopes is $_2^4He > \,_3^7Li > \,_4^9Be$. Therefore,the Assertion is incorrect.
Furthermore,the binding energy per nucleon does not increase linearly with the difference in the number of neutrons and protons; it depends on the mass defect and the stability of the nuclear configuration. Therefore,the Reason is also incorrect.