Question

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. $1.453\times {10}^{17}kg/m^3$

Answer 1

Answer: (D)

$1.453\times {10}^{17}kg/m^3$

Mass of the nucleus $M_N=A\times 1.67\times {10}^{-27}kg$ where A is the mass number of nucleus.
$R_0=1.4\times {10}^{-15}m$

Radius of nucleus $R=R_0{A}^{1/3}$
Thus volume of nucleus $V_N=\frac{4}{3}\pi R^3=\frac{4}{3}\pi (R_0{)}^{3}A$
The nuclear density is:

${\rho }_N=\frac{Massofnucleus}{Nuclearvolume}$

${\rho }_N=\frac{A\times 1.67\times {10}^{-27}}{\frac{4}{3}\pi (R_0{)}^{3}A}$
${\rho }_N=\frac{3\times 1.67\times {10}^{-27}}{12.56\times (1.4\times {10}^{-15}{)}^3}$
${\rho }_N=\frac{5.01\times {10}^{-27}}{34.46464\times {10}^{-45}}=1.453\times {10}^{17}kg{m}^{-3}$