Answer to Question #90141 in Physical Chemistry for Saudagar

Question

HCl molecule shows an absorption at 2886 cm^-1. Determine its force constant and maximum displacements (change in inter nuclear distance) for v=0,1,2,3

Answer 1

Solution.

$$\mu = \frac{m(H) \times m(Cl)}{m(H) + m(Cl)}$$

$$\mu = \frac{1 \times 35.5}{1 + 35.5}$$

$$\mu = 0.97 a.m.u.$$

$$\mu = 0.97 \times 1.66*10^{-27}$$

$$\mu = 1.61*10^{-27} kg$$

$$K = \mu \times (\frac{2 \times \pi \times c}{\lambda})^2$$

$$\lambda = \frac{1}{2886} * 10^{-7} = 3465 nm$$

$$K = 1.61*10^{-27} * (\frac{2*3.14*3*10^8}{3465*10^{-9}})^2$$

$$K = 475.97$$

$$E=\frac{h*c}{\lambda}$$

$$J = \frac{h^2}{\Delta E}$$

$$J = 1.21*10^{-48}$$

$$d = \sqrt{(\frac{J}{\mu})}$$

$$d = 2.74*10^{-11}$$

Answer:

K = 475.97 N/m

d = 2.74*10^(-11) m