Question

For a certain diatomic molecule, the lowest-energy photon observed in the vibrational spectrum is $\mathbf{0}.\mathbf{17}\mathbf{eV}$. What is the energy of a photon emitted in a transition from the $5$th excited vibrational energy level to the $2$nd excited vibrational energy level, assuming no change in the rotational energy?

Short answer

$1.02\text{eV}$

Step-by-step solution

Identification of the given data

The given data can be listed below as-

The energy of the photon at the ground state is,$E_0=0.17\text{eV}$.

Significance of the Planck’s law for finding the energy

The energy of the photon in the ground state can be determined by taking half of the product of Planck’s constant and wavelength of the photon. It is expressed as follows,

$E_0=\left(\frac{1}{2}\right)h{\omega }_0$

Here, $h$is the Planck’s constant and ${\omega }_0$is the wavelength of the photon.

Determination of the energy of the photon

The equation of the photon’s energy can be expressed as,

$$\begin{gathered} E_n=n+E_0 \\ =n+\left(\frac{1}{2}\right)h{\omega }_0 \end{gathered}$$

Here, $n$is the number of states, $E_0$is the energy of the photon at the ground state that is half of the product of the Planck’s constant $(h)$and wavelength of the photon $\left({\omega }_0\right)$$({\omega }_0)$.

Substitute localid="1662469232427" $n=2$for the second excited state in the above expression.

localid="1662469239412" $E_2=2+$localid="1662469245507" $\left(\frac{1}{2}\right)h{\omega }_0$

localid="1662469252499" $\frac{5}{2}h{\omega }_0$

Substitute localid="1662469258038" $n=5$for the fifth excited state in the above expression.

localid="1662469265782" $$\begin{gathered} E_5=5+\left(\frac{1}{2}\right)h{\omega }_0 \\ =\left(\frac{11}{2}\right)h{\omega }_0 \end{gathered}$$

Determination of the transition of the photon

The equation of the change in the energy transition of the photon can be expressed as,

$\Delta E=E_5-E_2$

Here, $E_5$is the energy of the fifth excited state and $E_2$is the energy of the second excited state.

For ,$E_2=\left(\frac{5}{2}\right)h{\omega }_0$and $E_5=\left(\frac{11}{2}\right)h{\omega }_0$and $\left(\frac{1}{2}\right)h{\omega }_0=0.17eV$

$$\begin{gathered} ∆E=\left(\frac{11}{2}\right)h{\omega }_0-\left(\frac{5}{2}\right)h{\omega }_0 \\ =6\times \left(\frac{1}{2}\right)h{\omega }_0 \\ =6\times 0.17eV \\ =1.02eV \end{gathered}$$

Thus, the energy of a photon emitted in a transition is $1.02\text{eV}$.