A) 0.5
B) 0.05
C) 0
D) 1.0
Correct Answer: B
Solution :
[b] \[1e=1.602\times {{10}^{-19}}C\] \[1\,esu=3.33\times {{10}^{-10}}C\] \[\frac{1e}{1\,esu}=\frac{1.602\times {{10}^{-19}}C}{3.33\times {{10}^{-10}}C}\] \[1e=4.802\times {{10}^{-10}}esu\] \[Dipole\text{ }moment=q\times distance\] \[\Rightarrow 1\,D\approx {{10}^{-18}}esucm\] \[0.38\times {{10}^{-18}}esu\,cm=q\times (1.61\times {{10}^{-18}}cm)\] \[q=2.36\times {{10}^{-11}}esu\] \[q=\frac{2.36\times {{10}^{-11}}esu}{4.802\times {{10}^{-10}}esu}\] \[q=0.049\] \[q\approx 0.05\]fractional charge
You need to login to perform this action.
You will be redirected in
3 sec
