Question

To a good approximation. the hydrogen chloride molecule, HCI, behaves vibrationally as a quantum harmonic ascillator of spring constant 480N/mand with effective osciltating mass just that of the lighter atom, hydrogen If it were in its ground vibtational state, what wave. Iength photon would be just right to bump this molecule. up to its next-higher vibrational energy state?.

Short answer

The wavelength$\lambda =3.51.{10}^{-6}m$

Step-by-step solution

Given Data

$k=480\frac{N}{m}$

Concept of the wavelength

In order to obtain the wavelength, need to calculate for the change in energy.$∆E$between E₀ and E₁ of the harmonic oscillator.

$\begin{array}{rcl}{E}_n& =& \left(n+\frac{1}{2}\right)\sout{h}\omega \\ ∆E& =& E_1-E_0\\ & =& \left(1+\frac{1}{2}\right)\sout{h}\omega -\left(0+\frac{1}{2}\right)\sout{h}\omega \\ & =& \sout{h}\omega \end{array}$

Where $∆E$ is change in energy between E₀ and E₁ of the harmonic oscillator, h is the plack’s constant and $\omega$is the wavelength.

Calculate the value of the wavelength by using the formula

The angular frequency, $\omega$, to the mass and spring constant:

$\begin{array}{rcl}\omega & =& \sqrt{\frac{k}{m}}∆E\\ & =& \sout{h}\sqrt{\frac{k}{m}}\\ & =& \left(1.055·{10}^{-34}·\left(\sqrt{\frac{480}{1.67·{10}^{-27}}}\right)\right)\\ & =& 5.65·{10}^{-20}J\end{array}$

Evaluate the value of the wavelength

Now the change in energy, $∆E$, as:

$\begin{array}{rcl}∆E& =& \frac{hc}{\lambda }\\ \lambda & =& \frac{hc}{∆E}\\ & =& \frac{\left(6.62·{10}^{-34}\right)\left(3·{10}^8\right)}{5.65·{10}^{-20}}\\ \lambda & =& 3.15·{10}^{-6}m\end{array}$

Thus, the wavelength$\begin{array}{rcl}\lambda & =& 3.15·{10}^{-6}m\end{array}$