Question

the force constant of CO molecule is 1860 Nm inverse . calculate vibrational frequency in cm inverse.​

Answer 1

Explanation:

If you notice, cm−1⋅cm/s=s −1 , as required for the units of ν0 . So, we obtain the fundamental vibrational frequency in the ...

I got, for 12C16O: • k=1856.92 N/m • E0=12hν0=2.129×10−20J • ΔE=hν0=4.258×10−20J We're treating the system as an anharmonic oscillator, .

Answer 2

Answer:

The vibrational frequency in cm⁻¹ is 2.16 × 10⁵ cm⁻¹

Explanation:

Force constant of CO molecule, k = 1860 N/m

Vibrational frequency, v =?

As we know,

• The vibrational frequency of a given molecule in terms of force constant can be calculated by the equation given below:
• $v = \frac{1}{2\pi } \sqrt{\frac{k}{\mu} }$

Here,

• v = vibrational frequency
• k = force constant
• μ = reduced mass

Now, we have to calculate the reduced mass of the CO molecule;

Reduced mass is given by,

• $\mu = \frac{m_{C} \times m_{0} }{m_{C} + m_{0}}$

Here,

• $m_{C}$ = mass of carbon in kg = 1.99 × 10⁻²⁶ kg
• $m_{O}$ = mass of oxygen in kg= 2.65 × 10⁻²⁶ kg

Therefore,

• $\mu = \frac{ (1.99 \times 10^{-26}) \times ( 2.65 \times 10^{-26}) }{(1.99 \times 10^{-26}) + ( 2.65 \times 10^{-26})}$
• $\mu = \frac{(1.99 \times 10^{-26}) \times ( 2.65 \times 10^{-26}) }{ 4.64 \times10^{-26})}$
• μ = 1.13 × 10⁻²⁶ kg

Now, vibrational frequency,

• $v = \frac{1}{2\pi } \sqrt{\frac{k}{\mu} }$
• $v = \frac{1}{2\times 3.14 } \sqrt{\frac{1860}{1.13\times 10^{-26} } }$
• $v = \frac{43.12}{6.28 \times 1.06 \times 10^{-13} }$

• v = 6.48 × 10¹³ s⁻¹  or 6.48 × 10¹³ Hz

Convert Hz into cm⁻¹

We know that,

• $v = \frac{c}{\lambda}$            
• $\frac{1}{\lambda} = \frac{v}{c}$
• $wavenumber = \frac{v}{c}$                 ( $wavenumber = \frac{1}{\lambda}$  )

Here,

• v = frequency
• c = speed of light = 3 × 10⁸ cm s
• λ = wavelength

Wavenumber is a unit of frequency in cm⁻¹.

• v in cm = $\frac{6.48 \times 10^{13} }{3 \times 10^{8} }$
• v = 2.16 × 10⁵ cm⁻¹

Hence, Vibrational frequency, v = 2.16 × 10⁵ cm⁻¹.