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# The density of a nucleus in which mass of each nucleon is $1.67 \times {10^{ - 27}}kg$ and ${R_0} = 1.4 \times {10^{ - 15}}m$is:(A) $2.995 \times {10^{17}}kg/{m^3}$(B) $1.453 \times {10^{16}}kg/{m^3}$(C) $1.453 \times {10^{19}}kg/{m^3}$(D) $1.453 \times {10^{17}}kg/{m^3}$

Last updated date: 20th Jun 2024
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Hint In the question mass of a single nucleon is given. We can use this to find the total mass of the nucleus. We know the formula for the radius of the nucleus. With the given values we can calculate the mass and volume of the nucleus. From this density can be calculated.
Formula used:
$\rho = \dfrac{m}{V}$ Where $\rho$stands for the density of the nucleus, $m$ stands for the mass of the nucleus, and $V$ stands for the volume of the nucleus.

Complete Step by step solution
Let $A$ be the number of nucleons in the nucleus.
Mass of each nucleon is given as, $1.67 \times {10^{ - 27}}kg$
Then the mass of the nucleus can be written as,
$m = A \times 1.67 \times {10^{ - 27}}kg$
The radius of a nucleus can be written as,
$R = {R_0}{A^{\dfrac{1}{3}}}$
It is given that, ${R_0} = 1.4 \times {10^{ - 15}}m$
Since the density of the nucleus can be written as $\rho = \dfrac{m}{V}$ we have to find the volume of the nucleus.
The nucleus is assumed to have a spherical shape. Therefore the volume of the nucleus can be written as,
$V = \dfrac{4}{3}\pi {R^3}$
Substituting $R = {R_0}{A^{\dfrac{1}{3}}}$ in the above equation, we get
$V = \dfrac{4}{3}\pi {\left( {{R_0}{A^{\dfrac{1}{3}}}} \right)^3}$
The volume of the nucleus is thus,
$V = \dfrac{4}{3}\pi R_0^3A$
The mass of the nucleus is,
$m = A \times 1.67 \times {10^{ - 27}}kg$
Now, we can write the density of the nucleus using the above values,
$\rho = \dfrac{m}{V} = \dfrac{{1.67 \times {{10}^{ - 27}}A}}{{\dfrac{4}{3}\pi R_0^3A}}$
$A$ will get cancelled and we substitute the value of ${R_0}$, we get
$\rho = \dfrac{{3 \times 1.67 \times {{10}^{ - 27}}}}{{4 \times 3.14 \times {{\left( {1.4 \times {{10}^{ - 15}}} \right)}^3}}} = 1.453 \times {10^{17}}$
The density of the nucleus will be $1.453 \times {10^{17}}kg/{m^3}$.
Option (D): $1.453 \times {10^{17}}kg/{m^3}$