Question

The length of the dipole of $HF$ molecule is $4\times {10}^{-11}m$. Calculate the dipole moment in coulomb-metres and in debyes.

• A. $5.5\times {10}^{-30}C-m$
• B. $6.4\times {10}^{-30}C-m$
• C. $7.4\times {10}^{-30}C-m$
• D. None of these

Answer 1

The correct option is B $6.4\times {10}^{-30}C-m$

$l=length=4\times {10}^{-11}m=4\times {10}^{-9}cm$
$e=charge=4.8\times {10}^{-10}esu$.
$\therefore \mu =e\times l=4.8\times {10}^{-10}\times 4\times {10}^{-9}esu-cm$
$=1.92\times {10}^{-18}esu-cm=1.92D$
since, $1D=3.3\times {10}^{-30}C-m$
$\mu =1.92\times 3.33\times {10}^{-30}C-m$
$\therefore =6.4\times {10}^{-30}C-m$