# How does the force constant change when a heavier atom is used in vibrational spectroscopy?

(iii) The fundamental band in the IR spectrum of $$\ce{^1H^{127}I}$$ is found at $$\pu{2308 cm^-1}.$$ Calculate the force constant for the bond. Assume the molecule can be treated as a harmonic oscillator.

(iv) Calculate how the position of the fundamental band would be expected to differ for molecules of $$\ce{^2H^{127}I}.$$

I have found the force constant in (iii), but I am struggling with (iv). I don’t know how to calculate the difference. Any insight would be greatly appreciated.

• You can calculate the ratio of the two frequencies as you know all the masses. Usually homework questions such as this get closed unless you explain what you have done so far and where you are stuck. As a start I suggest that you look a the vibrational spectroscopy part of your textbook where you are bound to find the equations you need. Apr 1 at 14:42

In the harmonic oscillator approximation, the wavenumber of vibration for a diatomic molecule is $$\tilde{\nu} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}$$ Here, $$k$$ is the force constant of your bond, and $$\mu$$ is the reduced mass for the two atoms. $$\mu=\frac{m_1m_2}{m_1+m_2}$$ When you change isotopes, only $$\mu$$ changes. The force constant depends on the strength of the bond, which depends on the chemical nature of the elements, and can be assumed to stay constant regardless of changing isotopes.
You have to calculate the $$\mu$$ for the second problem, and use that with the $$k$$ you got from the first problem to calculate the wavenumber of the fundamental band.
• Can the factor $k$ be obtained by some calculation ? Apr 1 at 12:31
As suggested by @porpyrin, in the second calculation you're assuming the force constant $$k$$ is constant in both molecules so a ratio can be taken. Let $$\tilde{\nu_H}$$ and $$\tilde{\nu_D}$$ be the hydrogen and deuterated versions respectively,
$$\tilde{\nu} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}$$ $$\frac{\tilde{\nu_H}}{\tilde{\nu_D}} = \frac{\sqrt{k/\mu_H}}{\sqrt{k/\mu_D}} = \sqrt{\frac{\mu_D}{\mu_H}}$$