Chapter 21 Problem 63

Problem

The HCl molecule has a dipole moment of about $3.4\times 10^{-30}{\textrm {C}}\cdot {\textrm {m}}$. The two atoms are separated by about $1.0\times 10^{-10}{\textrm {m}}$.

(a) What is the net charge on each atom?

(b) Is this equal to an integral multiple of $e$? If not, explain.

(c) What maximum torque would this dipole experience in a $2.5\times 10^{4}{\textrm {N/C}}$ electric field?

(d) How much energy would be needed to rotate one molecule 45° from its equilibrium position of lowest potential energy?

Solution

(a)

$p=Ql\rightarrow Q={\frac {p}{l}}={\frac {3.4\times 10^{-30}{\textrm {C}}\cdot {\textrm {m}}}{1.0\times 10^{-10}{\textrm {m}}}}=3.4\times 10^{-20}{\textrm {C}}$

(b)

${\frac {Q}{e}}={\frac {3.4\times 10^{-20}{\textrm {C}}}{1.6\times 10^{-19}{\textrm {C}}}}=0.21$ This is not a integer multiple of e. It means it is more probable to find the electron around Cl rather than H by a factor of 0.21 (or the wavefunction of the electron favors Cl side)

(c)

$\tau =pE\sin \theta \rightarrow \tau _{\textrm {max}}=pE=8.5\times 10^{-26}{\textrm {N}}\cdot {\textrm {m}}$

(d)

$W=\Delta U=\left(-pE\cos \theta _{\textrm {final}}\right)-\left(-pE\cos \theta _{\textrm {initial}}\right)$

$=pE\left(\cos \theta _{\textrm {initial}}-\cos \theta _{\textrm {final}}\right)$

$=8.5\times 10^{-26}{\textrm {N}}\cdot {\textrm {m}}\left(1-\cos 45^{\circ }\right)=2.5\times 10^{-26}{\textrm {J}}$