Answer:
Vibrational Energy Of HCl in the lowest state :
![2.86 \times 10^{-20} J](https://tex.z-dn.net/?f=2.86%20%5Ctimes%2010%5E%7B-20%7D%20J)
Classical Limit for stretching of HCl bond from its equilibrium length :
![Q_{0} = 0.0109 nm](https://tex.z-dn.net/?f=Q_%7B0%7D%20%3D%200.0109%20nm)
Percent of equilibrium Bond Length :
8.58 %
Explanation:
H-Cl bond Length = 0.127 nm =![1.27 \times 10^{-10}m](https://tex.z-dn.net/?f=%201.27%20%5Ctimes%2010%5E%7B-10%7Dm)
<u>Frequency </u>from<em> v = </em>0 to <em>v </em>= 1 is 2886
![\nu(Hz) = c\times \nu (cm^{-1})](https://tex.z-dn.net/?f=%5Cnu%28Hz%29%20%3D%20c%5Ctimes%20%5Cnu%20%28cm%5E%7B-1%7D%29)
![\nu(Hz) = 3 \times 10^{10} \times 2886](https://tex.z-dn.net/?f=%5Cnu%28Hz%29%20%3D%203%20%5Ctimes%2010%5E%7B10%7D%20%5Ctimes%202886%20)
![\nu = 8.568 \times 10^{13} Hz](https://tex.z-dn.net/?f=%5Cnu%20%3D%208.568%20%5Ctimes%2010%5E%7B13%7D%20Hz)
<u>Reduced mass</u>
![\mu =\frac{m_{1}m_{2}}{m_{1}+m_{2}}](https://tex.z-dn.net/?f=%5Cmu%20%3D%5Cfrac%7Bm_%7B1%7Dm_%7B2%7D%7D%7Bm_%7B1%7D%2Bm_%7B2%7D%7D)
![\mu =\frac{1 \times 35.5}{1+35.5}](https://tex.z-dn.net/?f=%5Cmu%20%3D%5Cfrac%7B1%20%5Ctimes%2035.5%7D%7B1%2B35.5%7D)
![\mu =\frac{1 \times 35.5}{36.5}](https://tex.z-dn.net/?f=%5Cmu%20%3D%5Cfrac%7B1%20%5Ctimes%2035.5%7D%7B36.5%7D)
but this has units in amu , to convert it in Kg divide it by
and 1000(to convert gram itno Kg)
<u>Calculation of Force constant :</u>
![\nu =\frac{1}{2\Pi }\sqrt{\frac{k}{\mu }}](https://tex.z-dn.net/?f=%5Cnu%20%3D%5Cfrac%7B1%7D%7B2%5CPi%20%7D%5Csqrt%7B%5Cfrac%7Bk%7D%7B%5Cmu%20%7D%7D)
Here,
![\nu = frequency](https://tex.z-dn.net/?f=%5Cnu%20%3D%20frequency)
k = force constant
![\mu = Reduced mass](https://tex.z-dn.net/?f=%5Cmu%20%3D%20Reduced%20mass)
Put the value of frequency , reduced mass and calculate for force constant
![2(3.14)\times 8.658\times 10^{13} = \sqrt{\frac{k}{1.61\times 10^{-27}}}](https://tex.z-dn.net/?f=2%283.14%29%5Ctimes%208.658%5Ctimes%2010%5E%7B13%7D%20%3D%20%5Csqrt%7B%5Cfrac%7Bk%7D%7B1.61%5Ctimes%2010%5E%7B-27%7D%7D%7D)
Solve the left hand side and square it. Then multiply it with reduced mass
k = 475.97 N/m
![\omega = 2(3.14)(8.568 \times 10^{13} Hz)](https://tex.z-dn.net/?f=%5Comega%20%3D%202%283.14%29%288.568%20%5Ctimes%2010%5E%7B13%7D%20Hz%29)
![\omega = 5.43 \times 10^{14}](https://tex.z-dn.net/?f=%5Comega%20%3D%205.43%20%5Ctimes%2010%5E%7B14%7D)
Calculation of lowest energy
![E_{0} = \frac{h\omega }{2\Pi }](https://tex.z-dn.net/?f=E_%7B0%7D%20%3D%20%5Cfrac%7Bh%5Comega%20%7D%7B2%5CPi%20%7D)
h = planck's constant =
![E_{0} = \frac{(6.626 \times 10^{-34})(5.43 \times 10^{14})}{2\Pi }](https://tex.z-dn.net/?f=E_%7B0%7D%20%3D%20%5Cfrac%7B%286.626%20%5Ctimes%2010%5E%7B-34%7D%29%285.43%20%5Ctimes%2010%5E%7B14%7D%29%7D%7B2%5CPi%20%7D)
On solving ,
<u>
</u>
Calculation of Stretching of HCl bond:
<em>Use formula :</em>
![\frac{1}{4\Pi }h\omega =\frac{1}{2}kQ_{0}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%5CPi%20%7Dh%5Comega%20%3D%5Cfrac%7B1%7D%7B2%7DkQ_%7B0%7D)
here Q = stretching bond length
Put , k = 475.97 N/m ,
and solve for Q
![Q_{0}^{2} = 1.204 \times 10^{-22}](https://tex.z-dn.net/?f=Q_%7B0%7D%5E%7B2%7D%20%3D%201.204%20%5Ctimes%2010%5E%7B-22%7D)
take square root
<u>
</u>
Calculation of Percentage extension:
Percentage![=\frac{Q_{0}}{X_{eq}}\times 100](https://tex.z-dn.net/?f=%20%3D%5Cfrac%7BQ_%7B0%7D%7D%7BX_%7Beq%7D%7D%5Ctimes%20100)
![\frac{1.097 \times 10^{-11}}{1.27 \times 10^{-10}}](https://tex.z-dn.net/?f=%5Cfrac%7B1.097%20%5Ctimes%2010%5E%7B-11%7D%7D%7B1.27%20%5Ctimes%2010%5E%7B-10%7D%7D)
Percentage = 8.58 %
<u />
<em> </em>