Atoms And Nuclei Question 166
Question: Order of magnitude of density of uranium nucleus is $ (m_{p}=1.67\times {{10}^{-27}}kg) $ [MP PET 1995; IIT-JEE 1999; MP PMT 2000; UPSEAT 2003]
Options:
A) $ 10^{20}kg/m^{3} $
B) $ 10^{17}kg/m^{3} $
C) $ 10^{14}kg/m^{3} $
D) $ 10^{11}kg/m^{3} $
Show Answer
Answer:
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} $ .
