A) \[{{10}^{20}}kg/{{m}^{3}}\]
B) \[{{10}^{17}}kg/{{m}^{3}}\]
C) \[{{10}^{14}}kg/{{m}^{3}}\]
D) \[{{10}^{11}}kg/{{m}^{3}}\]
Correct Answer: B
Solution :
The order of magnitude of mass and volume of uranium nucleus will be m ≃ A(1.67 ´ 10-27 kg) (A is atomic number) \[V=\frac{4}{3}\pi {{r}^{3}}\tilde{-}\frac{4}{3}\pi \ {{[(1.25\times {{10}^{-15}}m){{A}^{1/3}}]}^{3}}\] \[\tilde{-}(8.2\times {{10}^{-45}}{{m}^{3}})A\] Hence, \[\rho =\frac{m}{V}=\frac{A(1.67\times {{10}^{-27}}kg)}{(8.2\times {{10}^{-45}}{{m}^{3}})A}\] \[\tilde{-}2.0\times {{10}^{17}}kg/{{m}^{3}}\].
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