Question:
Order of magnitude of density of uranium nucleus is
$ (m_p = 1.67 \times 10^{-27} Kg ) $
Order of magnitude of density of uranium nucleus is
$ (m_p = 1.67 \times 10^{-27} Kg ) $
Updated On: Aug 15, 2022
- $ 10^{20} Kg/m^3 $
- $ 10^{17} Kg/m^3 $
- $ 10^{14} Kg/m^3 $
- $ 10^{11} Kg/m^3 $
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The Correct Option is B
Solution and Explanation
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 $
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In the year 1911, Rutherford discovered the atomic nucleus along with his associates. It is already known that every atom is manufactured of positive charge and mass in the form of a nucleus that is concentrated at the center of the atom. More than 99.9% of the mass of an atom is located in the nucleus. Additionally, the size of the atom is of the order of 10-10 m and that of the nucleus is of the order of 10-15 m.
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Following are the terms related to nucleus:
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