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Question: The nuclear radius of \(_8^{16}O\) is \(3 \times {10^{ - 15}}m\). If an atomic mass unit is \[1.67 \...

The nuclear radius of 816O_8^{16}O is 3×10−15m3 \times {10^{ - 15}}m. If an atomic mass unit is 1.67×10−271.67 \times {10^{ - 27}}kg, then the nuclear density is approximately:
a. 2.35×1017gcm−32.35 \times {10^{17}}gc{m^{ - 3}}
b. 2.35×1017kgm−32.35 \times {10^{17}}kg{m^{ - 3}}
c. 2.35×1017gm−32.35 \times {10^{17}}g{m^{ - 3}}
d. 2.35×1017kgmm−32.35 \times {10^{17}}kgm{m^{ - 3}}

Explanation

Solution

Density is known as the mass to volume ratio of some object. Assuming the nucleus to be spherical, find out its volume and hence the density.

Formula Used:
Density of the nucleus is given by:
D=MVD = \dfrac{M}{V} --(1)
Where,
D is the density,
M is the mass,
V is the volume.

Volume of a spherical object is given by:
V=43πR3V = \dfrac{4}{3}\pi {R^3} ----(2)
Where,
R is the radius of the object.

Complete answer:
Given:
The nucleus is of 816O_8^{16}O atom, which has a mass number 16, i.e. its mass is M=16amuM = 16amu.
One atomic mass unit (amu) is the same as 1.67×10−27kg1.67 \times {10^{ - 27}}kg.
Radius of the nucleus is R=3×10−15mR = 3 \times {10^{ - 15}}m.

To find: Density (D) of the nucleus.

Step 1
Using the eq.(2) in eq.(1) you’ll get the form:
D=M43πR3 ⇒D=3M4πR3  D = \dfrac{M}{{\tfrac{4}{3}\pi {R^3}}} \\\ \Rightarrow D = \dfrac{{3M}}{{4\pi {R^3}}} \\\ ----(3)

Step 2
Now, substitute the values of M and R in eq.(3) to get:
D=3×16amu4π×(3×10−15m)3 =3×16×1.67×10−274π×27×10−45kgm−3 =2.35×1017kgm−3  D = \dfrac{{3 \times 16amu}}{{4\pi \times {{\left( {3 \times {{10}^{ - 15}}m} \right)}^3}}} \\\ = \dfrac{{3 \times 16 \times 1.67 \times {{10}^{ - 27}}}}{{4\pi \times 27 \times {{10}^{ - 45}}}}kg{m^{ - 3}} \\\ = 2.35 \times {10^{17}}kg{m^{ - 3}} \\\

Hence, The density of nucleus is approximately (b) 2.35×1017kgm−32.35 \times {10^{17}}kg{m^{ - 3}}.

Note: When any atom is written in proper atomic notation ( like 816O_8^{16}O) the upper left number denotes the mass number i.e. total number of electron and protons where the lower left number is its atomic number i.e. the number of proton present in it. Total mass of the nucleus is the mass number in the atomic mass unit.
Also, from Bohr model of an atom you know that the maximum portion of atom remains empty where almost all mass of the atom remains at the nucleus of it in a very tiny space. This is the reason why nucleus is so much dense (notice the value again if you already haven’t).