A) 4m
B) 4 mm
C) \[4\mu \]m
D) 4 pm
Correct Answer: D
Solution :
There are 10 electrons and 10 protons in a neutral water molecule. So, its dipole moment is \[p=q(2I)=10e\,(2I)\] Hence length of the dipole ie, distance between centres of positive and negative charges is \[2l=\frac{p}{10e}=\frac{6.4\times {{10}^{-30}}}{10\times 1.6\times {{10}^{-19}}}\] \[=4\times {{10}^{-12}}m\] \[=4pm\]You need to login to perform this action.
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