
The radius of nucleus is:
A. proportional to its mass number
B. inversely proportional to its mass number
C. proportional to the cube root of its mass number
D. not related to its mass number
Answer
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Hint: Atomic nucleus is a small and dense region consisting of protons and neutrons at the center of an atom discovered by Ernest Rutherford in 1911. The size of the radius of the nucleus is 1-10 fm or femtometre.
Complete step by step answer:
Maximum mass of an atom is concentrated in the nucleus with negligible contribution from electrons. Nuclear force binds protons and neutrons to form a nucleus. Nuclear radius (R) is one of the basic quantities. For stable nucleus, the nuclear radius is approximately proportional to the cube root of the mass number (A) of the nucleus and the nucleons arranged in more spherical configurations.
Mathematically, represented as $\text{R = }{{\text{R}}_{0}}{{\text{A}}^{{1}/{3}\;}}$, where R is the radius and A is the atomic mass number (the number of protons plus the number of neutrons) and ${{\text{R}}_{0}}$is $1.25\times {{10}^{-15}}$m. The stable nucleus has approximately a constant density.
The correct answer to this question is option ‘c’ that the radius of the nucleus is proportional to the cube root of its mass number.
Note: The stability of an atom is counted by the stability of its nucleus. More stable a nucleus is the required energy is more per nucleon to pull the nucleus apart. The stability is caused by the attractive nuclear force between the nucleons. $_{56}\text{Fe}$ is the most stable nucleus. Repulsion between distant protons leads to instability and less binding energy per particle.
Complete step by step answer:
Maximum mass of an atom is concentrated in the nucleus with negligible contribution from electrons. Nuclear force binds protons and neutrons to form a nucleus. Nuclear radius (R) is one of the basic quantities. For stable nucleus, the nuclear radius is approximately proportional to the cube root of the mass number (A) of the nucleus and the nucleons arranged in more spherical configurations.
Mathematically, represented as $\text{R = }{{\text{R}}_{0}}{{\text{A}}^{{1}/{3}\;}}$, where R is the radius and A is the atomic mass number (the number of protons plus the number of neutrons) and ${{\text{R}}_{0}}$is $1.25\times {{10}^{-15}}$m. The stable nucleus has approximately a constant density.
The correct answer to this question is option ‘c’ that the radius of the nucleus is proportional to the cube root of its mass number.
Note: The stability of an atom is counted by the stability of its nucleus. More stable a nucleus is the required energy is more per nucleon to pull the nucleus apart. The stability is caused by the attractive nuclear force between the nucleons. $_{56}\text{Fe}$ is the most stable nucleus. Repulsion between distant protons leads to instability and less binding energy per particle.
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