Answer
When r=100 pm,
$\mu=1.6\times10^{-29}Cm$
When r= 200 pm,
$\mu=3.2\times10^{-29}Cm$
Work Step by Step
Dipole moment($\mu$)= charge(Q)$\times$distance of separation(r).
The magnitude of the charge on a proton or electron= $1.6\times10^{-19} C.$
When r=100 pm,
$\mu= 1.6\times10^{-19} C\times100 \times10^{-12}m=1.6\times10^{-29}Cm$
When r= 200 pm,
$\mu= 1.6\times10^{-19} C\times200 \times10^{-12}m=3.2\times10^{-29}Cm$
