Question

Given the mass of the iron nucleus as 55.85 u and A = 56, find the nuclear density.

Answer 1

Hint: Using Rutherford’s relation, the radius of the nucleus can be found. Further, the obtained should be multiplied with the mass of the electron to obtain the value of the mass. Then this mass value can be used to find the required value of the nuclear density of the iron.

Formula used:
\[\rho =\dfrac{m}{V}\]
\[r'={{r}_{0}}\times {{A}^{{}^{1}/{}_{3}}}\]
\[V=\dfrac{4}{3}\pi {{(r')}^{3}}\]

Complete step by step answer:
From given, we have the data,
The mass of the iron nucleus = 55.85 u
The mass number of the iron is, A = 56
According to Rutherford’s relation, the radius of the nucleus is given by the formula,
\[r'={{r}_{0}}\times {{A}^{{}^{1}/{}_{3}}}\]
Where \[{{r}_{0}}=1.2\times {{10}^{-15}}m\]and A is the mass number of the nucleus
Substitute the given values in the above equation to find the value of the nucleus of the iron.
So, we get,
\[\begin{align}
  & {{r}_{0}}=1.2\times {{10}^{-15}}m \\
 & r'=1.2\times {{10}^{-15}}\times {{(56)}^{{}^{1}/{}_{3}}} \\
 & r'=1.2\times {{10}^{-15}}\times 3.8258 \\
 & r'=4.59\times {{10}^{-15}}m \\
\end{align}\]
Therefore, the value of the nucleus of the iron is \[4.59\times {{10}^{-15}}m\].

Now compute the value of the mass of the iron nucleus.
We know that, \[1\,U=1.67\times {{10}^{-27}}kg\]
Therefore, the mass of the iron nucleus is,
\[\begin{align}
  & =55.85\times 1.67\times {{10}^{-27}}kg \\
 & =9.327\times {{10}^{-26}}kg \\
\end{align}\]

Finally, compute the nuclear density of the iron nucleus.
The density is given by the formula,
\[\rho =\dfrac{m}{V}\]
Where m is the mass and V is the volume
The volume is given by the formula,
\[V=\dfrac{4}{3}\pi {{r}^{3}}\]
Where r is the radius of the nucleus.

Substitute the values in the above equation to find the value of the density of the iron nucleus.
Therefore, we get,
\[\begin{align}
  & \rho =\dfrac{m}{V} \\
 & \rho =\dfrac{m}{{}^{4}/{}_{3}\pi {{(r')}^{3}}} \\
 & \rho =\dfrac{9.372\times {{10}^{-26}}}{{}^{4}/{}_{3}\times {}^{22}/{}_{7}\times {{(4.59\times {{10}^{-15}})}^{3}}} \\
 & \rho =2.3\times {{10}^{17}}{kg}/{{{m}^{3}}}\; \\
\end{align}\]

The nuclear density of the iron nucleus is \[2.29\times {{10}^{17}}{kg}/{{{m}^{3}}}\;\].

Note:
The things to be on your finger-tips for further information on solving these types of problems are: In this problem, the volume of the sphere should be used, as the nucleus will be in a spherical shape.