A) 0
B) 0.05
C) 0.5
D) 1.0
Correct Answer: B
Solution :
dipole moment \[\left( \mu \right)=q\times d\] d (distance) \[=1.617{\AA}=1.617\times {{10}^{-8}}cm\] \[\mu =0.38D=0.38\times {{10}^{-18}}esu\times cm\] \[q=\frac{\mu }{d}=\frac{0.38\times {{10}^{-18}}}{1.617\times {{10}^{-8}}}\] So fractional charge \[=\frac{\text{Particle}\,\text{chagre}}{\text{Total}\,\text{charge}}\] \[=\frac{q}{Q}=\frac{0.38\times {{10}^{-18}}}{1.617\times {{10}^{-8}}\times 4.802\times {{10}^{-10}}}=0.05\]
You need to login to perform this action.
You will be redirected in
3 sec
