Correct answer option is (A)

To determine the difference between the total mass of all the nucleons and the nuclear mass of the given nucleus, we need to use the concept of mass-energy equivalence provided by Einstein's famous equation:

$E=\mathrm{\Delta }m\cdot c^2$

where:

• E is the binding energy.
• $\mathrm{\Delta }m$ is the mass defect (difference in mass).
• $c$ is the speed of light in vacuum, which is approximately $3\times {10}^8\text{m/s}$.

Given:

• Binding energy, $E=18\times {10}^8\text{J}$.

We need to find $\mathrm{\Delta }m$, so rearranging the equation:

$\mathrm{\Delta }m=\frac{E}{c^2}$

Substitute the given values:

$\mathrm{\Delta }m=\frac{18\times {10}^8}{(3\times {10}^8{)}^2}$

Calculate the value:

$\mathrm{\Delta }m=\frac{18\times {10}^8}{9\times {10}^{16}}$

$\mathrm{\Delta }m=\frac{18}{9}\times {10}^{-8}$

$\mathrm{\Delta }m=2\times {10}^{-8}\text{kg}$

To convert the mass from kilograms to micrograms ($\mu g$), we use the conversion factor $1\text{kg}={10}^9\mu g$:

$\mathrm{\Delta }m=2\times {10}^{-8}\text{kg}\times {10}^9\frac{\mu g}{\text{kg}}$

$\mathrm{\Delta }m=2\times {10}^1\mu g$

$\mathrm{\Delta }m=20\mu g$

Therefore, the difference between the total mass of all the nucleons and the nuclear mass of the given nucleus is 20 $\mu g$.

Option A (20 $\mu g$) is the correct answer.

Correct answer option is (D)

In order to identify the correct nuclear fragments resulting from the fission of uranium-235 by a neutron, $\left({}_0^1\mathrm{n}\right)+\left({}_{92}^{235}\mathrm{U}\right)$, we need to apply the conservation of mass number and atomic number. These conservation laws tell us that the sum of mass numbers (top numbers, representing the total count of protons and neutrons) and the sum of atomic numbers (bottom numbers, representing the total count of protons) before and after the fission must be equal. Let's apply these rules to each option:

For the uranium-235 fission reaction, before the reaction, the total mass number is $235+1=236$ and the total atomic number is $92$ (since a neutron doesn't contribute to the atomic number).

Option A: ${}_{56}^{140}\mathrm{Xe}+{}_{38}^{94}\mathrm{Sr}+3{}_0^1\mathrm{n}$

• Total mass number after = $140+94+3(1)=234+3=237$
• Total atomic number after = $56+38=94$

Option B: ${}_{51}^{153}\mathrm{Sb}+{}_{41}^{99}\mathrm{Nb}+3{}_0^1\mathrm{n}$

• Total mass number after = $153+99+3(1)=252+3=255$
• Total atomic number after = $51+41=92$

Option C: ${}_{56}^{144}\mathrm{Ba}+{}_{36}^{89}\mathrm{Kr}+4{}_0^1\mathrm{n}$

• Total mass number after = $144+89+4(1)=233+4=237$
• Total atomic number after = $56+36=92$

Option D: ${}_{56}^{144}\mathrm{Ba}+{}_{36}^{89}\mathrm{Kr}+3{}_0^1\mathrm{n}$

• Total mass number after = $144+89+3(1)=233+3=236$
• Total atomic number after = $56+36=92$

By examining the conservation of mass numbers and atomic numbers, Option D is identified as the correct option because both the total mass number and atomic number after the reaction match exactly with the total mass number and atomic number before the reaction. Therefore, the nuclear fission fragments and the neutron count for the reaction are accurately represented by ${}_{56}^{144}\mathrm{Ba}+{}_{36}^{89}\mathrm{Kr}+3{}_0^1\mathrm{n}$.

Correct answer option is (C)

$\begin{array}{rl}& \begin{array}{rl}& (\mathrm{X})\to (\mathrm{Y})+(\mathrm{Z})+(\mathrm{P})\\ & \begin{array}{llll}\mathrm{M}& \mathrm{M}/3& \mathrm{M}/3& \mathrm{M}/3\end{array}\end{array}\\ & \mathrm{\Delta }{\mathrm{Mc}}^2=\frac{1}{2}\frac{\mathrm{M}}{3}{\mathrm{V}}^2+\frac{1}{2}\frac{\mathrm{M}}{3}{\mathrm{V}}^2+\frac{1}{2}\frac{\mathrm{M}}{3}{\mathrm{V}}^2\\ & \mathrm{V}=\mathrm{c}\sqrt{\frac{2\mathrm{\Delta }\mathrm{M}}{\mathrm{M}}}\end{array}$

Correct answer option is (A)

$\begin{array}{rl}& {}_3{\mathrm{Li}}^6+{}_0{\mathrm{n}}^1\to {}_2{\mathrm{He}}^4+{}_1{\mathrm{H}}^3\\ & {}_1{\mathrm{H}}^2+{}_1{\mathrm{H}}^3\to {}_2{\mathrm{He}}^4+{}_0{\mathrm{n}}^1\\ & {}_3{\mathrm{Li}}^6+{}_1{\mathrm{H}}^2\to 2\left({}_2{\mathrm{He}}^4\right)\\ & \end{array}$

Energy released in process

$\begin{array}{rl}& \mathrm{Q}=\mathrm{\Delta }{\mathrm{mc}}^2\\ & \mathrm{Q}=[\mathrm{M}(\mathrm{Li})+\mathrm{M}\left({\mathrm{H}}^2\right)-2\times \mathrm{M}\left({}_2{\mathrm{He}}^4\right)]\times 931.5\mathrm{MeV}\\ & \mathrm{Q}=[6.01690+2.01471-2\times 4.00388]\times 931.5\mathrm{MeV}\\ & \mathrm{Q}=22.216\mathrm{MeV}\\ & \mathrm{Q}=22.22\mathrm{MeV}\end{array}$

Correct answer option is (C)

$$ \begin{array}{rl}& {}_6{\mathrm{C}}^{13}+\text{ Energy }\to {}_6{\mathrm{C}}^{12}+{}_0{\mathrm{n}}^1\\ & \mathrm{\Delta }\mathrm{m}=(12.000000+1.008665)-13.003354\\ & =-0.00531\mathrm{u}\\ & \therefore \text{ Energy required }=0.00531\times 931.5\mathrm{MeV}\\ & =4.95\mathrm{MeV}\end{array}$$