A) \[1.5\times {{10}^{-8}}m\]
B) \[1.7\times {{10}^{-9}}m\]
C) \[1.5\times {{10}^{-10}}m\]
D) \[1.5\times {{10}^{-12}}m\]
Correct Answer: C
Solution :
Volume occupied by one gram atom of gold \[=\frac{197g}{19.7\,g/{{m}^{3}}}=10\,c{{m}^{3}}\] Volume of one atom\[=\frac{10}{6\times {{10}^{-23}}}\] \[=\frac{5}{3}\times {{10}^{23}}\,c{{m}^{3}}\] Let r be the radius of the atom \[\therefore \] \[\frac{4}{3}\pi {{r}^{3}}=\frac{5}{3}\times {{10}^{23}}\] \[{{r}^{3}}=\frac{50\times {{10}^{-24}}}{4\times 3.14}\] \[r=1.5\times {{10}^{-10}}\,m\]
You need to login to perform this action.
You will be redirected in
3 sec
