Question

Pleeease please help!!!!

P7E.4 The force constant for the bond in CO is 1857Nm-1. Calculate the vibrational frequencies (in Hz) of"C"О 13C"О "CO, and 13C180. Use integer relative atomic masses for this estimate.

Answer 1

Considering these as simple harmonic oscillators the equation to obtain the vibrational frequency can be written as:

$\nu = (1/2\pi )\sqrt{(K/\mu )}$ .................... Eqn 1

Where $\nu$ is the vibrational frequency in Sec^-1 or Hz, K is the force constant in dyne.Cm^-1, $\mu$ is the reduced mass in grams.

Reduces mass is given as:

$\mu = (m_{1}m_{2})/(m_{1}+m_{2})$ ....................... Eqn 2

Or, $\mu = (M_{1}M_{2})/(M_{1}+M_{2})$

Here M₁ and M₂ are the atomic mass in amu and m₁ and m₂ is the mass of the atoms involved in the bond in grams.

Its given that K = 1857 N.m^-1 = (1857 $\times$ 10⁵) / 100 dyne.Cm^-1 as 1 N = 10⁵ dyne and 1 m = 100 Cm.

So, K = 1857000 dyne.Cm^-1

Now Lets calculate the reduced mass using Eqn 2:

For ¹²C¹⁶O:

$\mu$ = (12 $\times$ 16)/ (12 + 16) = 6.857 amu = 6.857 / (6.023 $\times$ 10²³) g = 1.138 $\times$ 10^-23g

For ¹³C¹⁶O:

$\mu$ = (13 $\times$ 16)/ (13 + 16) = 7.1724 amu = 7.1724 / (6.023 $\times$ 10²³) g = 1.1908 $\times$ 10^-23g

For ¹²C¹⁸O:

$\mu$ = (12 $\times$ 18)/ (12 + 18) = 7.2 amu = 7.2 / (6.023 $\times$ 10²³) g = 1.1954 $\times$ 10^-23g

For ¹³C¹⁸O:

$\mu$ = (13 $\times$ 18)/ (13 + 18) = 7.548 amu = 7.548 / (6.023 $\times$ 10²³) g = 1.2532 $\times$ 10^-23g

Now substituting these values in Eqn 1 we have:

For ¹²C¹⁶O:

$\nu = (1/2\pi )\sqrt{(1857000 dyne.Cm^{-1}/1.138\times 10^{-23}g)}$

Or, $\nu = (1/2\pi )\sqrt{(1857000 g.Cm.S^{-2}.Cm^{-1}/1.138\times 10^{-23}g)}=6.4324\times 10^{13}S^{-1}orHz$

By substituting the values similarly we will get For ¹³C¹⁶O   $\nu$ = 6.2882 $\times$ 10¹³ Hz,

For ¹²C¹⁸O $\nu$ = 6.2760 $\times$ 10¹³ Hz and For ¹³C¹⁸O   $\nu$ = 6.1296 $\times$ 10¹³ Hz