Atoms & Nuclei

JEE Main level, single correct, medium

200 of 239

Question 200 of 239

For a nucleus XZAXZA having mass number A and atomic number ZZ
A. The surface energy per nucleon
(bs)=–a1A23.({b}_{s})=–{a}_{1}{A}^{\dfrac{2}{3}}.
B. The Coulomb contribution to the binding energy
bc=−a2Z(Z−1)A43{b}_{c}=-{a}_{2}\dfrac{Z(Z-1)}{{A}^{\dfrac{4}{3}}}.
C. The volume energy
bv=a3A{b}_{v}={a}_{3}A
D. Decrease in the binding energy is proportional to surface area.

E. While estimating the surface energy, it is assumed that each nucleon interacts with 12 nucleons. (
a1,a2{a}_{1},{a}_{2} and a3{a}_{3} are constants)
Choose the most appropriate answer from the options given below:

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Answer: (B) C,DC,D only

The formula to calculate the radius of a nucleus is given by
R=R0A13...(1)R={R}_{0}{A}^{\dfrac{1}{3}}...\left(1\right)
Hence, the volume energy can be written as

bv=C1R3=a3(A13)3=a3A\begin{matrix}{b}_{v} & = & {C}_{1}{R}^{3} \\ & = & {a}_{3}{\left({A}^{\dfrac{1}{3}}\right)}^{3} \\ & = & {a}_{3}A\end{matrix}
The nuclear binding energy is the minimum energy that is required to disassemble the nucleus of an atom into its constituent protons and neutrons. Also, the energy of the nucleus is negative with regard to the energy of the particles pulled apart to infinite distance, because energy must be utilised to split a nucleus into its individual protons and neutrons. It is proportional to the surface area.

Hence, statements C and D are most appropriate.

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