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Q.
Order of magnitude of density of uranium nucleus is
$ (m_p = 1.67 \times 10^{-27} Kg ) $
IIT JEEIIT JEE 1999
Solution:
Radius of a nucleus is given by
$ R = R_0 A^{1/3} (Where R_0 = 1.25 \times 10^{-15} m ) $
$ = 1.25 A^{1/3} \times 10^{-15} m $
Here A is the mass number and mass of the uranium nucleus
will be
$ m = Am_p $ where $ m_p = mass \, of \, proton $
$ = A (1.67 \times 10^{-27} Kg ) $
$ \therefore $ Density $ p = \frac{mass}{volume} = \frac{m}{\frac{4}{3} \pi R^3} $
$ = \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 $