Nucleus (Stability and Reaction) Questions in English

(B) An $\alpha$-particle corresponds to a helium nucleus $({}_{2}^{4}He)$. It reduces the mass number by $4$ and the atomic number by $2$.
In $\beta$-decay,the atomic number increases by $1$ while the mass number remains unchanged.
For the reaction: ${}_{90}^{232}Th \rightarrow {}_{82}^{208}Pb + x({}_{2}^{4}He) + y({}_{-1}^{0}e)$.
Equating the mass numbers: $232 = 208 + 4x$,which gives $4x = 24$,so $x = 6$.
Equating the atomic numbers: $90 = 82 + 2x - y$.
Substituting $x = 6$: $90 = 82 + 12 - y$,which gives $90 = 94 - y$,so $y = 4$.
Thus,$6 \alpha$ and $4 \beta$ particles are emitted.

(B) The binding energy $(BE)$ is calculated using the mass defect $(\Delta m)$:
$\Delta m = [Z \times M_{H} N \times M_{n}] - M_{nucleus}$
Here,$Z = 3$ (protons),$N = 4$ (neutrons),$M_{H} = 1.007825 \ u$,$M_{n} = 1.008665 \ u$,and $M_{Li} = 7.016005 \ u$.
$\Delta m = [3 \times 1.007825 4 \times 1.008665] - 7.016005$
$\Delta m = [3.023475 4.034660] - 7.016005 = 7.058135 - 7.016005 = 0.04213 \ u$
$BE = \Delta m \times 931 \ MeV/u = 0.04213 \times 931 \approx 39.2 \ MeV$.

(A) The nuclear reaction is: $_{90}Th^{228}$ $\xrightarrow{-4 \alpha} _{82}Pb^{212}$ $\xrightarrow{-1 \beta} _{83}Bi^{212}$.
The daughter element is $_{83}Bi^{212}$.
The number of neutrons is calculated as: $\text{Mass number} - \text{Atomic number}$.
$\text{Number of neutrons} = 212 - 83 = 129$.

(C) For the occurrence of nuclear fusion, a very high temperature (i.e., $20$ million $K$ or $2 \times 10^{7} \,K$) is required.
Thus, these reactions are also known as thermonuclear reactions.

(A) The binding energy $(BE)$ released is related to the mass defect $(\Delta m)$ by the equation: $BE = \Delta m \times 931 \ MeV$.
Given $BE = 306.3 \ MeV$,we calculate the mass defect:
$\Delta m = \frac{306.3}{931} = 0.3290 \ u$.
The mass defect is defined as: $\Delta m = \text{Calculated mass} - \text{Isotopic mass}$.
Therefore,$\text{Isotopic mass} = \text{Calculated mass} - \Delta m$.
$\text{Isotopic mass} = 40.328 - 0.3290 = 39.998 \ u$.

(A) The nuclear reaction is: ${}^{238}_{92}U \longrightarrow {}^{206}_{82}Pb + m({}^{4}_{2}He) + n({}^{0}_{-1}e)$.
Comparing the mass numbers on both sides: $238 = 206 + 4m$ $\Rightarrow 4m = 32$ $\Rightarrow m = 8$.
Comparing the atomic numbers on both sides: $92 = 82 + 2m - n$.
Substituting $m = 8$: $92 = 82 + 2(8) - n$ $\Rightarrow 92 = 82 + 16 - n$ $\Rightarrow 92 = 98 - n$ $\Rightarrow n = 6$.
Therefore,$8$ $\alpha$-particles and $6$ $\beta$-particles are emitted.

(A) Let the reaction be ${}_{90}Th^{232} \rightarrow {}_{82}Pb^{208} + n_{\alpha} ({}_{2}He^{4}) + n_{\beta} ({}_{-1}e^{0})$.
Mass balance: $232 = 208 + 4n_{\alpha}$ $\Rightarrow 4n_{\alpha} = 24$ $\Rightarrow n_{\alpha} = 6$.
Atomic number balance: $90 = 82 + 2n_{\alpha} - n_{\beta}$ $\Rightarrow 90 = 82 + 2(6) - n_{\beta}$ $\Rightarrow 90 = 94 - n_{\beta}$ $\Rightarrow n_{\beta} = 4$.
Therefore,$6 \alpha$ and $4 \beta$ particles are emitted.

(C) $Uranium$ and $plutonium$ are commonly used as fuel in an atomic pile (nuclear reactor).

(A) $Cleveite$ is a variety of uraninite and it contains uranium,which is a radioactive element.

(B) Given,mass defect $\Delta m = 0.081 \ u$.
Total number of nucleons $A = 11$.
Binding energy $BE = \Delta m \times 931 \ MeV/u = 0.081 \times 931 = 75.411 \ MeV$.
Average binding energy per nucleon $= \frac{BE}{A} = \frac{75.411}{11} = 6.855 \ MeV \approx 6.85 \ MeV$.

(B) Given,mass defect $\Delta m = 0.081 \ u$.
Number of nucleons $A = 11$.
Binding energy $= 931 \times \Delta m \ MeV = 931 \times 0.081 \ MeV = 75.411 \ MeV$.
Average binding energy $= \frac{\text{Binding energy}}{A} = \frac{75.411}{11} \ MeV = 6.85 \ MeV$.

(D) $1$. The radius of a nucleus is given by $R = R_0 A^{1/3}$,where $A$ is the mass number. Option $A$ is incorrect because it states $A^{1/2}$.
$2$. Isobars are atoms with the same mass number but different atomic numbers. $_{7}N^{15}$ and $_{8}O^{16}$ have different mass numbers ($15$ and $16$),so they are not isobars. Option $B$ is incorrect.
$3$. The thorium series ($4n$ series) ends at the stable isotope $_{82}Pb^{208}$. Option $C$ is incorrect as it mentions $_{83}Bi^{209}$.
$4$. Magic numbers for nucleons are $2, 8, 20, 28, 50, 82, 126$. For $_{20}Ca^{40}$,the number of protons is $20$ and the number of neutrons is $40 - 20 = 20$. Since both $20$ and $20$ are magic numbers,option $D$ is correct.

(B) The mass-energy equivalence is given by Einstein's equation $E = mc^2$.
For $1 \ a.m.u.$ (atomic mass unit),the energy equivalent is calculated as follows:
$1 \ a.m.u. = 1.6605 \times 10^{-27} \ kg$.
Using $c = 2.9979 \times 10^8 \ m/s$,the energy $E = (1.6605 \times 10^{-27} \ kg) \times (2.9979 \times 10^8 \ m/s)^2 \approx 1.4924 \times 10^{-10} \ J$.
Since $1 \ eV = 1.6022 \times 10^{-19} \ J$,then $1 \ MeV = 1.6022 \times 10^{-13} \ J$.
Dividing the energy in Joules by the conversion factor for $MeV$:
$E = \frac{1.4924 \times 10^{-10} \ J}{1.6022 \times 10^{-13} \ J/MeV} \approx 931.5 \ MeV$.
Since $1 \ MeV = 10^6 \ eV$,the energy is $931.5 \times 10^6 \ eV$.

(B) The mass defect is given as $3.32 \times 10^{-26} \ g$.
First,convert the mass defect into atomic mass units $(amu)$ by dividing by the mass of one $amu$ $(1.66 \times 10^{-24} \ g)$:
$\text{Mass defect in } amu = \frac{3.32 \times 10^{-26} \ g}{1.66 \times 10^{-24} \ g/amu} = 0.02 \ amu$.
The binding energy is calculated using the relation: $\text{Binding energy} = \text{Mass defect in } amu \times 931 \ MeV/amu$.
$\text{Binding energy} = 0.02 \times 931 \ MeV = 18.62 \ MeV$.

(A) When a positron is emitted,a proton is converted into a neutron,a positron,and a neutrino as shown in the reaction: $^1_1H \rightarrow ^1_0n + ^0_{+1}e + \nu$.
Since a proton is converted into a neutron,the total number of nucleons (mass number) remains constant,but the number of protons (atomic number) decreases by $1$ unit.

(B) Nuclei with a low $n/p$ ratio are proton-rich and unstable. They achieve stability by increasing the $n/p$ ratio. This can occur through two primary processes:
$1$. Positron emission: $^1_1p \rightarrow ^1_0n + ^0_{+1}e + \nu$
$2$. $K$-electron capture: $^1_1p + ^0_{-1}e \rightarrow ^1_0n + \nu$
Both processes convert a proton into a neutron,thereby decreasing the number of protons and increasing the number of neutrons,which increases the $n/p$ ratio. Since both options $(B)$ and $(C)$ are correct,in many contexts,positron emission is the primary answer provided for this specific question type.

(D) The radioactive decay of $_{11}Na^{24}$ occurs via $\beta^-$-decay.
In $\beta^-$-decay,a neutron is converted into a proton,increasing the atomic number by $1$ while the mass number remains constant.
The nuclear reaction is: $_{11}Na^{24} \longrightarrow _{12}Mg^{24} + _{-1}\beta^{0}$.

(C) The process described is electron capture,where the nucleus $^{64}_{29}Cu$ captures an orbital electron $(-1e^0)$.
In this process,the atomic number decreases by $1$ while the mass number remains constant.
The nuclear equation is: $^{64}_{29}Cu + _{-1}e^0 \longrightarrow ^{64}_{28}Ni$.
Thus,the product is $^{64}_{28}Ni$.

(D) Among the given options,$_{92}^{238}U$ is an isotope of uranium but cannot be used as a nuclear fuel.
$_{92}^{238}U$ is non-fissile,meaning it does not undergo fission by thermal neutrons.
The energy released when $_{92}^{238}U$ absorbs a neutron is insufficient to carry out nuclear fission.
Hence,$(d)$ is the correct option.

(C) Radioactivity is a nuclear phenomenon that depends solely on the composition and stability of the atomic nucleus.
It is independent of the chemical environment,such as the formation of compounds or the presence of orbital electrons.
Therefore,radium in its elemental form and radium in the form of radium chloride $(RaCl_2)$ exhibit the same level of radioactivity.