
The size of atomic nucleus is of the order of:
A. ${10^{ - 15}}\,m$
B. ${10^{10}}\,m$
C. ${10^{14}}\,m$
D. ${10^{10}}\,m$
Answer
566.7k+ views
Hint: Atomic nucleus consists of protons and neutrons and is surrounded by an electron cloud. The diameter of the nucleus is from $1.6\,fm$ to $15\,fm$. Neutrons are free of charge and protons are positively charged.
Complete step by step answer:
The nuclear core is the little, thick locale comprising of protons and neutrons at the focal point of a particle, found in $1911$ by Ernest Rutherford dependent on the $1909$ Geiger–Marsden gold foil activity. After the disclosure of the neutron in $1932$, models for a core made out of protons and neutrons were immediately evolved by Dmitri Ivanenko and Werner Heisenberg. A particle is made out of a decidedly charged core, with a haze of contrarily charged electrons encompassing it, bound together by electrostatic power. Practically the entirety of the mass of a molecule is situated in the core, with a contribution from the electron cloud. Protons and neutrons are bound together to shape a core by the atomic power.
The core of a molecule comprises neutrons and protons, which thus are the sign of more rudimentary particles, called quarks that are held by the strong nuclear power in certain steady composition of hadrons, called baryons. The atomic solid power stretches out far enough from every baryon to tie the neutrons and protons together against the repulsive electrical power between the positively charged protons. The atomic solid power has a short reach, and basically drops to zero just past the edge of the core. The aggregate activity of the decidedly charged core is to hold the electrically negative charged electrons in their circles about the core.
Size of the atomic nucleus can be determined as:
$R = {R_ \circ }{A^{\dfrac{1}{3}}}$
Where ${R_ \circ } = 1.5 \times {10^{ - 15}}\,m$ and $A$ which is the mass number is of the order of $1$.
Hence, $R = {10^{15}}\,m$
Thus option A is correct.
Note:Here we have taken the diameter of the heaviest atom to get the size of atomic number. Also, we have to take the order of mass numbers as one.
Complete step by step answer:
The nuclear core is the little, thick locale comprising of protons and neutrons at the focal point of a particle, found in $1911$ by Ernest Rutherford dependent on the $1909$ Geiger–Marsden gold foil activity. After the disclosure of the neutron in $1932$, models for a core made out of protons and neutrons were immediately evolved by Dmitri Ivanenko and Werner Heisenberg. A particle is made out of a decidedly charged core, with a haze of contrarily charged electrons encompassing it, bound together by electrostatic power. Practically the entirety of the mass of a molecule is situated in the core, with a contribution from the electron cloud. Protons and neutrons are bound together to shape a core by the atomic power.
The core of a molecule comprises neutrons and protons, which thus are the sign of more rudimentary particles, called quarks that are held by the strong nuclear power in certain steady composition of hadrons, called baryons. The atomic solid power stretches out far enough from every baryon to tie the neutrons and protons together against the repulsive electrical power between the positively charged protons. The atomic solid power has a short reach, and basically drops to zero just past the edge of the core. The aggregate activity of the decidedly charged core is to hold the electrically negative charged electrons in their circles about the core.
Size of the atomic nucleus can be determined as:
$R = {R_ \circ }{A^{\dfrac{1}{3}}}$
Where ${R_ \circ } = 1.5 \times {10^{ - 15}}\,m$ and $A$ which is the mass number is of the order of $1$.
Hence, $R = {10^{15}}\,m$
Thus option A is correct.
Note:Here we have taken the diameter of the heaviest atom to get the size of atomic number. Also, we have to take the order of mass numbers as one.
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