Question

The ratio of density of the nucleus of ${}_{47}^{108}\text{Ag}$ assuming $r_{nucleus}$ is $1.3A^{1/3}\times {10}^{-13}cm$, where $A$ is mass number of nucleus, with density of metallic silver $10.5gc{m}^3$, expressed in terms of $\times {10}^{13}$ is:

Answer 1

Radius of ${}_{47}^{108}\text{Ag}$ nucleus
$=r=[1.3\times (A{)}^{1/3}\times {10}^{-13}]cm=[1.3\times (108{)}^{1/3}\times {10}^{-13}]cm$
Here, A is the mass number of the nucleus.
Volume of ${}_{47}^{108}\text{Ag}$ nucleus
$=\frac{4}{3}\pi r^3=\frac{4}{3}\times \frac{22}{7}\times [1.3\times (108{)}^{1/3}\times {10}^{-13}cm{]}^3$
$=9.94\times {10}^{-37}c{m}^3$
The density of nucleus$=\frac{\text{mass}}{\text{volume}}$
Mass of nucleus $=\frac{\text{Molar mass}}{\text{avogadro's number}}=\frac{108}{6.023\times {10}^{23}}$
The density of nucleus $=\frac{108}{6.023\times {10}^{23}\times 9.94\times {10}^{-37}}=1.80\times {10}^{14}g/c{m}^3$
Thus, $\frac{\text{density of nucleus}}{\text{density of atom}}=\frac{1.80\times {10}^{14}}{10.5}$
$=1.72\times {10}^{13}$ = $2\times {10}^{13}$