Nucleus (Stability and Reaction) Questions in English

(A) The density of the nucleus is approximately $2.3 \times 10^{17} \ kg/m^3$.
This value is independent of the mass number $A$ because the volume of the nucleus is proportional to $A$ $(V \propto A)$,and the mass of the nucleus is also proportional to $A$ $(m \propto A)$.
Therefore,the density $\rho = \frac{m}{V}$ remains constant for all nuclei,which is of the order of $10^{17} \ kg/m^3$.

(B) As the atomic number $(Z)$ increases,the electrostatic Coulomb repulsion between protons increases significantly. To maintain nuclear stability,additional neutrons are required to increase the strong nuclear force (which is independent of charge) without adding further electrostatic repulsion. This effectively dilutes the proton-proton repulsion,allowing the nucleus to remain stable at higher mass numbers.

(A) . $A$ fusion reaction is difficult to occur because positively charged nuclei repel each other.
At very high temperatures of the order of $10^6$ to $10^7 \ K$,the nuclei may have sufficient energy to overcome the repulsive forces and fuse.
It is for this reason,fusion reactions are also called thermonuclear reactions.
Hence,hydrogen can be fused to form helium at high temperature and high pressure.

(C) Nuclear fusion is the process where two light nuclei combine to form a heavier nucleus. In the case of hydrogen isotopes,the reaction is: $1H^2 + 1H^2 \to 2He^4$. Thus,the product is Helium.

(A) Radioactive elements are inherently unstable due to an imbalance in the number of protons and neutrons in their nucleus.
This instability is characterized by a relatively low binding energy per nucleon compared to stable isotopes.
Therefore,the nucleus of a radioactive element possesses low binding energy,which leads to radioactive decay to achieve a more stable state.

(B) Nuclear energy is based on the principle of mass-energy equivalence,as described by Einstein's equation $E = mc^2$.
In nuclear reactions,a small amount of mass is lost,which is converted into a large amount of energy.
Therefore,the correct option is $(B)$.

(D) $Plutonium-239$ $(^{239}Pu)$ is a fissile material,meaning it can sustain a nuclear chain reaction when bombarded with thermal neutrons.
It is commonly used as fuel in nuclear reactors and in nuclear weapons due to its high probability of undergoing fission.

(C) nuclear reaction must be balanced in terms of mass and energy. In nuclear reactions,the total mass-energy is conserved,as described by Einstein's mass-energy equivalence principle,$E = mc^2$.

(C) To find the missing product,we balance the mass numbers and atomic numbers on both sides of the reaction.
Mass number balance: $130 + 2 = 131 + A \implies 132 = 131 + A \implies A = 1$.
Atomic number balance: $52 + 1 = 53 + Z \implies 53 = 53 + Z \implies Z = 0$.
The particle with mass number $1$ and atomic number $0$ is a neutron,represented as $_{0}n^{1}$.
Thus,the reaction is: $_{52}Te^{130} + _{1}H^{2} \to _{53}I^{131} + _{0}n^{1}$.

(C) In a nuclear reaction,the sum of atomic numbers and the sum of mass numbers must be conserved on both sides.
For the given reaction: $_{7}N^{14} + _{1}H^{1} \to _{8}O^{15} + X$
Sum of atomic numbers on the left side: $7 + 1 = 8$.
Sum of atomic numbers on the right side: $8 + Z = 8$,so $Z = 0$.
Sum of mass numbers on the left side: $14 + 1 = 15$.
Sum of mass numbers on the right side: $15 + A = 15$,so $A = 0$.
$A$ particle with atomic number $0$ and mass number $0$ is a gamma photon,denoted as $\gamma$.
Therefore,the correct option is $C$.

(C) Magic numbers are $2, 8, 20, 28, 50, 82,$ and $126$.
For $_{82}Pb^{208}$,the number of protons is $Z = 82$ and the number of neutrons is $N = A - Z = 208 - 82 = 126$.
Since both $82$ and $126$ are magic numbers,$_{82}Pb^{208}$ is a doubly magic nuclide,which imparts high stability to the nucleus.

(B) The carbon cycle (also known as the $CNO$ cycle) is a process of stellar nucleosynthesis in which stars on the main sequence fuse hydrogen into helium.
In this cycle,carbon nuclei act as catalysts.
They facilitate the fusion process but are not consumed in the net reaction.
Therefore,the carbon nucleus is regenerated at the end of the cycle.

(A) For lead $(_{82}Pb^{208})$: Number of neutrons $(n)$ = $208 - 82 = 126$. Number of protons $(p)$ = $82$. The ratio $n/p = \frac{126}{82} \approx 1.536$.
For bismuth $(_{83}Bi^{209})$: Number of neutrons $(n)$ = $209 - 83 = 126$. Number of protons $(p)$ = $83$. The ratio $n/p = \frac{126}{83} \approx 1.518$.
Comparing the two,$1.536 > 1.518$,therefore the $n/p$ ratio is higher for lead.

(A) The correct answer is $(A)$.
Positron is the anti-particle of an electron,having the same mass as an electron but with a positive charge $(+1)$.

(A) For light elements with low atomic numbers (up to $Z = 20$),stable nuclides typically have a neutron-to-proton ratio of approximately $1$ $(n/p = 1)$.
As the atomic number increases,the number of neutrons required for stability increases,leading to an $n/p$ ratio greater than $1$ for heavier elements.
However,in the context of general introductory chemistry,the ideal ratio for stability is often represented as $n/p = 1$.

(A) The formation of a nucleus from its constituent nucleons (protons and neutrons) is accompanied by the release of energy,known as binding energy.
According to Einstein's mass-energy equivalence principle,$E = \Delta m c^2$,this energy release corresponds to a decrease in the total mass of the system.
This difference in mass is known as the mass defect $(\Delta m)$,where the mass of the nucleus is less than the sum of the masses of its individual nucleons.

(B) positron is an antiparticle of the electron,having the same mass as an electron but a positive charge. It is represented as $_{+1}e^{0}$ or $\beta^{+}$.

(D) The stability of an atom is determined by the binding energy per nucleon and the ratio of neutrons to protons.
Among the given options,$Pb$ ($Lead$,atomic number $82$) is the final stable product of the radioactive decay series of heavy elements like $U$ $(Uranium)$ and $Th$ $(Thorium)$.
Therefore,$Pb$ is the most stable atom among the choices provided.

(A) The stability of a nucleus is influenced by the pairing of protons and neutrons.
Nuclei with an even number of protons $(X)$ and an even number of neutrons $(Y)$ are known as $Even-Even$ nuclei.
These $Even-Even$ nuclei are the most stable and abundant in nature due to the pairing effect,which leads to higher binding energy per nucleon.
Conversely,$Odd-Odd$ nuclei are generally the least stable.

(B) The packing fraction is defined as the difference between the isotopic mass and the mass number,divided by the mass number.
Mathematically,$f = \frac{M - A}{A}$.
$A$ lower (or more negative) value of the packing fraction indicates a higher binding energy per nucleon,which corresponds to greater nuclear stability.
Since atom $B$ has a lower packing fraction than atom $A$,atom $B$ is more stable than atom $A$.

(D) In a nuclear explosion, the energy is released primarily as $kinetic \ energy$ of the fission fragments, $\gamma \ radiation$, and $neutrons$. Since none of the options $A$, $B$, or $C$ fully describe the primary form of energy release (which is a combination of kinetic and electromagnetic energy), the correct choice is $D$.

(A) The meson was predicted and discovered by the Japanese physicist $Hideki \ Yukawa$ in $1935$ to explain the strong nuclear force between nucleons.

(C) When a heavy nucleus like $_{92}U^{235}$ is bombarded with slow neutrons,it splits into two lighter nuclei,releasing a large amount of energy and more neutrons. This process is known as nuclear fission.
The reaction is represented as: $_{92}U^{235} + _0n^1 \to _{56}Ba^{141} + _{36}Kr^{92} + 3_0n^1 + \text{Energy}$.

(C) The radioactive decay series of Uranium (specifically $U-238$) proceeds through a series of alpha and beta emissions until it reaches a stable isotope. The final stable product of the Uranium decay series is $Pb-206$ (Lead).

(A) The energy required to separate the nucleons (protons and neutrons) from a nucleus is known as the $Binding \ energy$.
Therefore,the correct option is $(A)$.

(C) The energy liberated in a nuclear reaction is calculated using the mass-energy equivalence relation: $E = \Delta m \times 931 \; MeV/amu$.
Given,loss of mass $\Delta m = 0.01864 \; amu$.
Therefore,energy liberated = $0.01864 \times 931 \; MeV$.
Energy liberated = $17.35584 \; MeV \approx 17.36 \; MeV$.

(A) The mass defect $(\Delta m)$ is calculated as the difference between the sum of the masses of the reactants and the sum of the masses of the products.
$\Delta m = (\text{mass of } _1H^2 + \text{mass of } _1H^3) - (\text{mass of } _2He^4 + \text{mass of } _0n^1)$
$\Delta m = (2.014 + 3.016) - (4.004 + 1.008) \ amu$
$\Delta m = 5.030 - 5.012 = 0.018 \ amu$

(B) The radioactive series are classified based on the remainder when the mass number $(A)$ is divided by $4$:
$1$. For $_{95}Am^{241}$,the mass number is $241$. Dividing $241$ by $4$ gives a remainder of $1$ $(241 = 4 \times 60 + 1)$,so it belongs to the $4n + 1$ series (Neptunium series).
$2$. For $_{90}Th^{234}$,the mass number is $234$. Dividing $234$ by $4$ gives a remainder of $2$ $(234 = 4 \times 58 + 2)$,so it belongs to the $4n + 2$ series (Uranium series).
Therefore,the correct option is $B$.

(D) Radioactivity is a characteristic property of an unstable nucleus. It occurs when an unstable atomic nucleus loses energy by radiation.

(C) The nuclear reaction is: $_{92}U^{238} \xrightarrow{-8\alpha, -6\beta} _{Z}X^{A}$.
Change in mass number $(A)$: $238 - (8 \times 4) = 238 - 32 = 206$.
Change in atomic number $(Z)$: $92 - (8 \times 2) + (6 \times 1) = 92 - 16 + 6 = 82$.
The product nucleus is $_{82}X^{206}$.
Number of protons $(p)$ $= 82$.
Number of neutrons $(n)$ $= A - Z = 206 - 82 = 124$.
Neutron/proton ratio $= n/p = 124/82 = 62/41$.

(A) The $\beta$-emission process is represented by the equation: $_{6}C^{14} \to _{7}N^{14} + _{-1}e^{0}$.
In this reaction,the parent nucleus is $C^{14}$.
The number of neutrons in the parent nucleus $C^{14}$ is calculated as: $\text{Mass number} - \text{Atomic number} = 14 - 6 = 8$.

(A) $\beta$ emission increases the atomic number by $1$ while the mass number remains unchanged.
Let the parent nucleus be $_Z X^A$.
After two successive $\beta$ emissions,the reaction is: $_Z X^A \xrightarrow{-2\beta} _{Z+2} N^{14}$.
Comparing the atomic number and mass number,we get $Z+2 = 7$,so $Z = 5$,and $A = 14$.
The parent nucleus is $_5 X^{14}$.
The number of neutrons in $_5 X^{14}$ is $A - Z = 14 - 5 = 9$.

(C) The end product of each natural radioactive series is a stable isotope of lead $(Pb)$.

(B) The atomic number $(Z)$ of $_{13}^{29}Al$ is $13$ and the number of neutrons $(n)$ is $29 - 13 = 16$.
The $\frac{n}{p}$ ratio is $\frac{16}{13} \approx 1.23$.
For light elements,the stable $\frac{n}{p}$ ratio is approximately $1$.
Since the $\frac{n}{p}$ ratio of $_{13}^{29}Al$ is higher than the stable ratio,it lies above the belt of stability.
To achieve stability,it converts a neutron into a proton by emitting a $\beta$-particle $(_{-1}^{0}e)$: $_{13}^{29}Al \rightarrow _{14}^{29}Si + _{-1}^{0}e$.

(B) In a nuclear reaction,both the total atomic number (sum of subscripts) and the total mass number (sum of superscripts) must be conserved on both sides of the equation.
For option $A$: $3+1 = 4$ (atomic) and $7+1 = 8 \neq 7+1 = 8$. (Correct)
For option $B$: $21+0 = 21$ and $20+0 = 20$. Since $21 \neq 20$,the atomic number is not conserved. (Incorrect)
For option $C$: $33+2 = 35$ and $75+4 = 79 \neq 78+1 = 79$. (Correct)
For option $D$: $83+1 = 84$ and $209+2 = 211 \neq 210+1 = 211$. (Correct)
Therefore,option $B$ is not a balanced nuclear reaction.

(B) The $(4n + 2)$ series is known as the Uranium series.
In this series,the mass number of the starting element is $238$,which satisfies the formula $4n + 2$ (where $n = 59$).
The end product of the Uranium series is a stable isotope of lead,which is $_{82}Pb^{206}$.
Therefore,the correct option is $B$.

(A) The thorium series is a $4n$ radioactive decay series.
The parent element of this series is $_{90}Th^{232}$,and it undergoes a series of $\alpha$ and $\beta$ decays to reach a stable end product.

$4n$ Series (Name)	Parent to End Element
$4n$: Thorium series $4n+1$: Neptunium series $4n+2$: Uranium series $4n+3$: Actinium series	$_{90}Th^{232} \rightarrow _{82}Pb^{208}$ $_{94}Pu^{241} \rightarrow _{83}Bi^{209}$ $_{92}U^{238} \rightarrow _{82}Pb^{206}$ $_{92}U^{235} \rightarrow _{82}Pb^{207}$

Therefore,the end product of the thorium series is $_{82}Pb^{208}$.

(A) Radium $(Ra)$ belongs to the uranium series,which is the $(4n + 2)$ series.
The radioactive disintegration process in this series terminates at a stable isotope of lead,specifically $^{206}Pb$.
Therefore,the end product of the disintegration of radium is lead.

(A) The $(4n + 1)$ series is known as the Neptunium series.
The starting member is $_{93}Np^{237}$ and the stable end product is $_{83}Bi^{209}$.

(A) The neptunium series is a radioactive decay series that starts with $Np-237$ and ends with the stable isotope $Bi-209$. Therefore,the radioactivity stops when the element is converted to $Bi$.

(A) The binding energy per nucleon is a measure of the stability of a nucleus.
Elements with mass numbers in the range of $40$ to $100$ are the most stable.
Among the given options,$Fe$ (Iron-$56$) has a mass number of $56$,which falls in the region of maximum stability.
Therefore,$Fe$ has the highest binding energy per nucleon.

(C) The initial isotope is $_{90}Th^{232}$.
Each $\alpha$-particle emission reduces the mass number by $4$ and the atomic number by $2$.
Each $\beta$-particle emission increases the atomic number by $1$ and leaves the mass number unchanged.
Final mass number $= 232 - (6 \times 4) = 232 - 24 = 208$.
Final atomic number $= 90 - (6 \times 2) + (4 \times 1) = 90 - 12 + 4 = 82$.
The element with atomic number $82$ is Lead $(Pb)$.
Therefore,the final isotope is $_{82}Pb^{208}$.

(D) The sequence of radioactive decays starting from a radioactive parent nucleus and ending at a stable daughter nucleus is known as a radioactive disintegration series or decay series.

(D) The $\beta$-emission process is represented as: $_{Z}^{A}X \rightarrow _{Z+1}^{A}Y + _{-1}^{0}e$.
Given that the daughter nucleus is $^{14}N$ $(_{7}^{14}N)$,the parent nucleus must have an atomic number $Z = 7 - 1 = 6$ and mass number $A = 14$.
Thus,the parent nucleus is $^{14}C$ $(_{6}^{14}C)$.
The number of neutrons in the parent nucleus is calculated as $A - Z = 14 - 6 = 8$.

(A) For $_{18}Ar^{40}$:
Number of protons $= 18$
Number of neutrons $= 40 - 18 = 22$
Mass defect $(\Delta m) = [18 \times m_{p} + 22 \times m_{n}] - \text{mass of } Ar$
$\Delta m = [18 \times 1.007825 + 22 \times 1.008665] - 39.962384$
$\Delta m = [18.14085 + 22.19063] - 39.962384$
$\Delta m = 40.33148 - 39.962384 = 0.369096 \, amu$
Binding energy $= \Delta m \times 931.5 \, MeV/amu$
Binding energy $= 0.369096 \times 931.5 \approx 343.81 \, MeV$.

(B) In the first reaction:
$_{92}^{238}M \to _{y}^{x}N + 2_{2}^{4}He$
Applying conservation of mass number: $x = 238 - (2 \times 4) = 230$
Applying conservation of atomic number: $y = 92 - (2 \times 2) = 88$
So,the element is $_{88}^{230}N$.
In the second reaction:
$_{88}^{230}N \to _{B}^{A}L + 2\beta^{+}$
For $\beta^{+}$ decay (positron emission),the mass number $A$ remains the same and the atomic number decreases by $1$ per particle.
$A = 230$
$B = 88 - (2 \times 1) = 86$
So,the element is $_{86}^{230}L$.
Number of neutrons in $_{86}^{230}L = A - B = 230 - 86 = 144$.

(A) The element has $6$ neutrons and $6$ protons (since it is Carbon-$12$,atomic mass $12.00710 \, amu$).
Mass of $6$ neutrons $= 6 \times 1.00893 \, amu = 6.05358 \, amu$.
Mass of $6$ protons $= 6 \times 1.00814 \, amu = 6.04884 \, amu$.
Total mass of nucleons $= 6.05358 + 6.04884 = 12.10242 \, amu$.
Mass defect $(\Delta m) = 12.10242 - 12.00710 = 0.09532 \, amu$.
Binding energy $= \Delta m \times 931.5 \, MeV = 0.09532 \times 931.5 = 88.7956 \, MeV$.
Binding energy per nucleon $= 88.7956 / 12 = 7.399 \, MeV \approx 7.39 \, MeV$.

(D) Step $1$: The beta decay of $_{35}X^{88}$ is represented as: $_{35}X^{88} \to _{36}W^{88} + _{-1}e^{0}$.
Step $2$: The resulting nucleus $_{36}W^{88}$ is unstable and emits a neutron $(_{0}n^{1})$: $_{36}W^{88} \to _{36}W^{87} + _{0}n^{1}$.
Step $3$: The final product is $_{36}W^{87}$.

(A) In radioactive $\beta^-$-decay,a neutron in the nucleus is converted into a proton and an electron (beta particle).
The reaction is: $_0n^1 \to _{+1}p^1 + _{-1}e^0$.
Therefore,the emitted electron originates from the nucleus of the atom.

(C) The binding energy per nucleon is calculated by dividing the total binding energy by the mass number $(A)$.
For $_8O^{16}$,the mass number $A = 16$.
Binding energy per nucleon $= \frac{127 \ MeV}{16} = 7.94 \ MeV$.