Answer:
Vibrational Energy Of HCl in the lowest state :
![2.86 * 10^(-20) J](https://img.qammunity.org/2020/formulas/chemistry/college/g002kvgol9lluct3jqmdg9nbyvrjyue4ss.png)
Classical Limit for stretching of HCl bond from its equilibrium length :
![Q_(0) = 0.0109 nm](https://img.qammunity.org/2020/formulas/chemistry/college/se338gwo5y56wx46fsu445ywe9y0eluqex.png)
Percent of equilibrium Bond Length :
8.58 %
Step-by-step explanation:
H-Cl bond Length = 0.127 nm =
![1.27 * 10^(-10)m](https://img.qammunity.org/2020/formulas/chemistry/college/15mz1u2jbm3fpgd5mnwjdoe8cti0subvdo.png)
Frequency from v = 0 to v = 1 is 2886
![\\u(Hz) = c* \\u (cm^(-1))](https://img.qammunity.org/2020/formulas/chemistry/college/3cuaiknmjtmcbhm6g9bbigvbhggg55k9qs.png)
![\\u(Hz) = 3 * 10^(10) * 2886](https://img.qammunity.org/2020/formulas/chemistry/college/v7ufrjpal7ouk63f5dm16vw6hphobti5pp.png)
![\\u = 8.568 * 10^(13) Hz](https://img.qammunity.org/2020/formulas/chemistry/college/qqdlqdxscyvgz8wjuor1lfep8z1y6ytz77.png)
Reduced mass
![\mu =(m_(1)m_(2))/(m_(1)+m_(2))](https://img.qammunity.org/2020/formulas/chemistry/college/2irmkn6lkujdy3mlxx4qosb24o7qqvlyaj.png)
![\mu =(1 * 35.5)/(1+35.5)](https://img.qammunity.org/2020/formulas/chemistry/college/7u3huj13wiptyktc7uhhtydzwr8uzud0o9.png)
![\mu =(1 * 35.5)/(36.5)](https://img.qammunity.org/2020/formulas/chemistry/college/m58nzuy3ohbvhcusjjsx3ta356u9mv704p.png)
but this has units in amu , to convert it in Kg divide it by
and 1000(to convert gram itno Kg)
Calculation of Force constant :
![\\u =(1)/(2\Pi )\sqrt{(k)/(\mu )}](https://img.qammunity.org/2020/formulas/chemistry/college/4zkh0428yis6fsmzqe0u2u2mp3zu89wzh5.png)
Here,
![\\u = frequency](https://img.qammunity.org/2020/formulas/physics/college/z643v03bcid5nq3swy07biynrlo1442rcl.png)
k = force constant
![\mu = Reduced mass](https://img.qammunity.org/2020/formulas/chemistry/college/y9ozh71yyejz7yhqwby48cdnu8ks2uucr7.png)
Put the value of frequency , reduced mass and calculate for force constant
![2(3.14)* 8.658* 10^(13) = \sqrt{(k)/(1.61* 10^(-27))}](https://img.qammunity.org/2020/formulas/chemistry/college/ztkbuxwe2vnn65zy5fuzvysrdaz3mnh2dw.png)
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 * 10^(13) Hz)](https://img.qammunity.org/2020/formulas/chemistry/college/buwcwe7dui278su91xmid6fi4smvz9rido.png)
![\omega = 5.43 * 10^(14)](https://img.qammunity.org/2020/formulas/chemistry/college/cyc5co1xrbrpkf00eyj20cj8uh9d0loeqk.png)
Calculation of lowest energy
![E_(0) = (h\omega )/(2\Pi )](https://img.qammunity.org/2020/formulas/chemistry/college/r4ru2laixh404kqef6e2jsjcmhc42108rs.png)
h = planck's constant =
![E_(0) = ((6.626 * 10^(-34))(5.43 * 10^(14)))/(2\Pi )](https://img.qammunity.org/2020/formulas/chemistry/college/i94j4hbinjwlkmobbamblegldnq10o5o89.png)
On solving ,
![E_(0) = 2.86 * 10^(-20)J](https://img.qammunity.org/2020/formulas/chemistry/college/9kv9h63is9w6b5zgwrcpro9o9nrgxnai12.png)
Calculation of Stretching of HCl bond:
Use formula :
![(1)/(4\Pi )h\omega =(1)/(2)kQ_(0)](https://img.qammunity.org/2020/formulas/chemistry/college/qyftgy0m578019hyjz66thz8q6zrqwm2ck.png)
here Q = stretching bond length
Put , k = 475.97 N/m ,
and solve for Q
![Q_(0)^(2) = 1.204 * 10^(-22)](https://img.qammunity.org/2020/formulas/chemistry/college/l166oo02fszlaho2xjdscxusaivjl9kk7i.png)
take square root
![Q_(0) = 1.097 * 10^(-11)](https://img.qammunity.org/2020/formulas/chemistry/college/pt9w1jugyy6t74geow29iv3sl2wehpd077.png)
Calculation of Percentage extension:
Percentage
![=(Q_(0))/(X_(eq))* 100](https://img.qammunity.org/2020/formulas/chemistry/college/27a2of0vzejvm7k2fv9ij0019lf788lprc.png)
![(1.097 * 10^(-11))/(1.27 * 10^(-10))](https://img.qammunity.org/2020/formulas/chemistry/college/al9dqane5z3pdzhvbbd9ndpzk3ch6wa32z.png)
Percentage = 8.58 %