Solveeit Logo
Login

Question

Physics Question on Nuclei

Order of magnitude of density of uranium nucleus is (mp=1.67×10−27Kg)(m_p = 1.67 \times 10^{-27} Kg )

A

1020Kg/m310^{20} Kg/m^3

B

1017Kg/m310^{17} Kg/m^3

C

1014Kg/m310^{14} Kg/m^3

D

1011Kg/m310^{11} Kg/m^3

Answer

1017Kg/m310^{17} Kg/m^3

Explanation

Solution

Radius of a nucleus is given by R=R0A1/3(WhereR0=1.25×10−15m)R = R_0 A^{1/3} (Where R_0 = 1.25 \times 10^{-15} m ) =1.25A1/3×10−15m= 1.25 A^{1/3} \times 10^{-15} m Here A is the mass number and mass of the uranium nucleus will be m=Ampm = Am_p where mp=mass of protonm_p = mass \, of \, proton =A(1.67×10−27Kg)= A (1.67 \times 10^{-27} Kg ) ∴\therefore Density p=massvolume=m43πR3p = \frac{mass}{volume} = \frac{m}{\frac{4}{3} \pi R^3} =A(1.67×10−27Kg)A(1.25×10−15m)3  or  p≈2.0×1017Kg/m3= \frac{A(1.67 \times 10^{-27} Kg)}{A(1.25 \times 10^{-15}m)^3} \, \, or \, \, p \approx 2.0 \times 10^{17} Kg/m^3