Question

Shown below is a portion of the rotational spectrum of CO. The main peaks nd to transitions of the c6O molecule, while the minor peaks correspond to other isotopic correspon combinations. Using data obtained from this spectrum, determine the equilibrium bond length re of 12c16O in Å. C12018 100 一、 90 80 70 60 50 40 30 20- 10 15 20 25 Wavenumber/cm1 30 35 40

Answer 1

5.

C13016 C12018 100 90 80 70 60 50 a 40 30 20 . 1아 2B- 4 cm 15 20 25 Wavenumber/cm1 30 35 40

Bond length,

$r_{e} =\sqrt{\frac{h}{8\pi ^{2}cB\mu }}$

where,

h = Planck's constant = 6.62 x 10^-34 J.s

c = speed of light in vacuum = 3 x 10⁸ m/s

B = rotational constant = 200 m^-1   (As, 2B = (distance between any two peaks) = 4 cm^-1 = 400 m^-1)

reduced mass,

$\mu = \frac{m_{1}m_{2}}{m_{1}+m_{2}}$

m₁ = mass of one C atom = 12 x 1.66 x 10^-24 g = 0.012 x 1.66 x 10^-24 kg

m₂ = mass of one O atom = 16 x 1.66 x 10^-24 g = 0.016 x 1.66 x 10^-24 kg

$\mu = \frac{0.012\times 1.66 \times 10^{-24}\times 0.016\times 1.66 \times 10^{-24}}{0.012\times 1.66 \times 10^{-24}+0.016\times 1.66 \times 10^{-24}}$

     = 0.0069 x 1.66 x 10^-24 kg

Hence,

$r_{e} =\sqrt{\frac{6.62\times 10^{-34}}{8 \times 3.14^{2} \times 3\times 10^{8}\times 200\times 0.0069\times 1.66\times 10^{-24}}}$

or, r_e = 1.11 x 10^-10 m

    = 1.11 $\aA$$\AA$

Therefore, the equilibrium bond length of ¹²C¹⁶O = 1.11 $\AA$

Answer 2

5.

C13016 C12018 100 90 80 70 60 50 a 40 30 20 . 1아 2B- 4 cm 15 20 25 Wavenumber/cm1 30 35 40

Bond length,

$r_{e} =\sqrt{\frac{h}{8\pi ^{2}cB\mu }}$

where,

h = Planck's constant = 6.62 x 10^-34 J.s

c = speed of light in vacuum = 3 x 10⁸ m/s

B = rotational constant = 200 m^-1   (As, 2B = (distance between any two peaks) = 4 cm^-1 = 400 m^-1)

reduced mass,

$\mu = \frac{m_{1}m_{2}}{m_{1}+m_{2}}$

m₁ = mass of one C atom = 12 x 1.66 x 10^-24 g = 0.012 x 1.66 x 10^-24 kg

m₂ = mass of one O atom = 16 x 1.66 x 10^-24 g = 0.016 x 1.66 x 10^-24 kg

$\mu = \frac{0.012\times 1.66 \times 10^{-24}\times 0.016\times 1.66 \times 10^{-24}}{0.012\times 1.66 \times 10^{-24}+0.016\times 1.66 \times 10^{-24}}$

     = 0.0069 x 1.66 x 10^-24 kg

Hence,

$r_{e} =\sqrt{\frac{6.62\times 10^{-34}}{8 \times 3.14^{2} \times 3\times 10^{8}\times 200\times 0.0069\times 1.66\times 10^{-24}}}$

or, r_e = 1.11 x 10^-10 m

    = 1.11 $\aA$$\AA$

Therefore, the equilibrium bond length of ¹²C¹⁶O = 1.11 $\AA$