Question

Order of magnitude of density of uranium nucleus is

• A. ${10}^{20}kg/m^3$
• B. ${10}^{17}kg/m^3$
• C. ${10}^{14}kg/m^3$
• D. ${10}^{11}kg/m^3$

Answer 1

Answer: (B)

${10}^{17}kg/m^3$

$\begin{array}{c}{R}_0=1.25\times {10}^{-15}m\\ A=\text{Mass no.}\\ V=\frac{4}{3}\pi R^3\\ m=A\times \text{mass of}\text{proton}=A\times 1.67\times {10}^{-27}kg\end{array}$

$\begin{array}{c}\text{Radius of nucleus is given}by\\ R=R_0{A}^{1/3}\\ \text{Density}=\frac{\text{Mass}}{\text{Volume}}\\ D=\frac{m}{V}\end{array}$

$=\frac{A\times 1.67\times {10}^{-27}}{A\times \frac{4}{3}\pi \times {(1.25\times {10}^{-15})}^3}$

$\begin{array}{cc}D=& 2\times {\mathrm{10}}^{17}cg{m}^{-3}\\ \text{order of magnitude}& ={10}^{17}kg/m^3\end{array}$