Answer
See the detailed answer below.
Work Step by Step
To calculate the first three vibrational energy levels of a $$\rm{C=O} $$
carbon-oxygen double bond using the data from the figure (which shows the absorption spectrum of acetone), we need to use the wavelength corresponding to the vibrational transition and the basic concepts of vibrational energy levels.
The molecule is described as a quantum harmonic oscillator, meaning its energy levels are given by
$$
E_n = \left(n - \frac{1}{2}\right)\hbar \omega\tag 1
$$
We look at the energy of the first three levels: $ E_1 $, $ E_2 $, and $ E_3 $.
The absorption occurs because a photon is absorbed, causing a transition between two energy levels (from $ n=1 $ to $ n=2 $). The energy of the absorbed photon is given by
$$
E_{\text{photon}} = hf = \frac{hc}{\lambda}\tag 2
$$
where, from the mentioned graph, we can see that $\lambda = 5.75 \; \rm{\mu m} $.
The difference in energy between the two levels is given by
$$
\Delta E = E_2 - E_1 =E_{\text{photon}}$$
$$
E_{\text{photon}}= \left(2 - \frac{1}{2}\right)\hbar \omega-\left(1 - \frac{1}{2}\right)\hbar \omega=\hbar\omega
$$
Plug from (2);
$$
\frac{hc}{\lambda}=\hbar\omega$$
where $\hbar=h/2\pi$
$$\frac{\color{red}{\bf\not} hc}{\lambda}= \dfrac{\color{red}{\bf\not} h}{2\pi}\omega
$$
Thus,
$$\omega=\frac{2\pi c}{\lambda}\tag 3$$
Plug into (1);
$$
E_n = \left(n - \frac{1}{2}\right)\frac{2\pi \hbar c}{\lambda} = \left(n -\frac{1}{2}\right)\frac{2\pi h c}{2\pi \lambda}$$
$$E_n = \dfrac{ \left(n - \frac{1}{2}\right) h c}{ \lambda}
$$
Thus,
$$
E_1 = \dfrac{ \left(1 - \frac{1}{2}\right) (6.63\times 10^{-34})(3\times 10^8)}{ (5.75\times 10^{-6})}=\bf 1.73\times10^{-20}\;\rm J\\\;\\\;\\
E_1 =\color{red}{\bf 0.108 }\;\rm eV
$$
$$
E_2 = \dfrac{ \left(2 - \frac{1}{2}\right) (6.63\times 10^{-34})(3\times 10^8)}{ (5.75\times 10^{-6})}=\bf 5.19\times10^{-20}\;\rm J\\\;\\\;\\
E_2 =\color{red}{\bf 0.324}\;\rm eV
$$
$$
E_3 = \dfrac{ \left(3 - \frac{1}{2}\right) (6.63\times 10^{-34})(3\times 10^8)}{ (5.75\times 10^{-6})}=\bf 8.65\times10^{-20}\;\rm J\\\;\\\;\\
E_3 =\color{red}{\bf 0.540}\;\rm eV
$$
