Question:

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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Concepts Used:

Nuclei

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.

Read More: Nuclei

Following are the terms related to nucleus:

  1. Atomic Number
  2. Mass Number
  3. Nuclear Size
  4. Nuclear Density
  5. Atomic Mass Unit