Question

Order of magnitude of density of uranium nucleus is $\left( {{m}_{p}}=1.67\times {{10}^{-27}}Kg \right)$.
$\begin{align}
  & \text{A}\text{. }{{10}^{20}}Kg{{m}^{-3}} \\
 & \text{B}\text{. }{{10}^{17}}Kg{{m}^{-3}} \\
 & \text{C}\text{. }{{10}^{14}}Kg{{m}^{-3}} \\
 & \text{D}\text{. }{{10}^{11}}Kg{{m}^{-3}} \\
\end{align}$

Answer 1

Hint: The order of magnitude of a quantity is an exponential change of plus or minus one in the value of that quantity or unit. We will calculate the density of the uranium nucleus using the formula density is the ratio of mass to the volume. For calculating the mass of the uranium nucleus, we will multiply the mass of a proton by the total number of protons present in a uranium nucleus. Volume of the uranium nucleus can be calculated using the relation between the radius of a heavy nucleus with the mass number of an atom.


Complete step by step answer:
Uranium is a chemical element being represented with the symbol U and has the atomic number equal to 92. It is a heavy metal which has been used as an abundant source of concentrated energy for more than 60 years. Uranium occurs in most rocks in concentrations of 2 to 4 parts per million and is commonly found in the Earth's crust as tin, tungsten and molybdenum. Uranium occurs in seawater, and can be recovered from the oceans as well. .
Like other elements, Uranium occurs in several slightly differing forms known as isotopes. These isotopes differ from each other in the number of uncharged particles that are neutrons, inside the nucleus. Natural uranium which is found in the Earth's crust is a mixture largely of two isotopes: uranium-238 (U-238), and uranium-235 (U-235).
We are given a uranium nucleus and we have to calculate the order of magnitude of density of its nucleus.
The order of magnitude is usually written as 10 to the ${{n}^{th}}$ power. The $n$ represents the order of magnitude of a particular unit or quantity. If we raise a number by one order of magnitude, it means we are basically multiplying that number by 10. If we are decreasing a number by one order of magnitude, we are basically dividing that number by 10.
Radius of a nucleus is given by,
$R={{R}_{o}}{{A}^{\dfrac{1}{3}}}$
Where,
${{R}_{o}}=1.25\times {{10}^{-15}}m$
And, $A$ is the mass number of an atom
Mass of an atom is given by,
$m=A{{m}_{p}}$
Where,
${{m}_{p}}$is the mass of proton
For uranium, we have,
$m=A\left( 1.67\times {{10}^{-15}} \right)kg$
Density is given as,
$\text{Density = }\dfrac{\text{Mass}}{\text{Volume}}$
Volume of spherical atom,
$V=\dfrac{4}{3}\pi {{R}^{3}}$
Therefore,
$D=\dfrac{m}{V}$
Putting values,
$\begin{align}
  & D=\dfrac{A\times 1.67\times {{10}^{-27}}}{A\times \dfrac{4}{3}\times \pi \times {{\left( 1.25\times {{10}^{-15}} \right)}^{3}}} \\
 & D=\dfrac{A\times 1.67\times {{10}^{-27}}}{A\times \dfrac{4}{3}\times \dfrac{22}{7}\times {{\left( 1.25\times {{10}^{-15}} \right)}^{3}}} \\
 & D=2\times {{10}^{17}}Kg{{m}^{-3}} \\
\end{align}$
Order of magnitude of density of uranium nucleus is ${{10}^{17}}Kg{{m}^{-3}}$

So, the correct answer is “Option B”.

Note:
While calculating the mass of an atom or a nucleus, we only take the mass of protons and neutrons present inside the nucleus. Since the electrons are considered to be massless, they have very less mass as compared to a proton or a neutron, their mass is not considered while determining the mass of a nucleus. This is why we say that all the mass of an atom resides at the centre of the atom inside the nucleus.