Chemistry: The mass of the deuterium molecule D2 is twice that of the hydrogen molecule H2. If the vibrational frequency of H2 is 1.29 × 1014 Hz, what is the vibrational frequency of D2, assuming that the spring constant of attracting forces is the same for the two species? Answer in units of Hz.

well they d have higher potential energyso gravitational...

The mass of the deuterium molecule D2 is twice that

Hannahmiller3773

26.01.2022

9.12x10¹³ Hz

Explanation:

The vibrational frequency (ω) of a molecule is given by:

$\omega = \sqrt{\frac{k}{\mu}}$

Where:

k: is the spring constant

μ: is the reduced mass

The reduced mass of a diatomic molecule is:

$\frac{1}{\mu} = \frac{1}{m_{a}} + \frac{1}{m_{b}}$

Where ma and mb are the atomic masses of the atoms a and b, respectively, of the diatomic molecule.

Hence, the vibrational frequency of the hydrogen molecule is:

$\omega_{H_{2}} = \sqrt{\frac{k}{\mu_{H_{2}}}}$ (1)

From equation (1) we can find k:

$k = \omega_{H_{2}}^{2}*\mu_{H_{2}}$ (2)

The vibrational frequency of the deuterium molecule is:

$\omega_{D_{2}} = \sqrt{\frac{k}{\mu_{D_{2}}}}$ (3)

By entering equation (2) into equation (3) we can calculate the vibrational frequency of the deuterium molecule:

$\omega_{D_{2}} = \sqrt{\frac{\omega_{H_{2}}^{2}*\mu_{H_{2}}}{\mu_{D_{2}}}}$

$\omega_{D_{2}} = \sqrt{\frac{\omega_{H_{2}}^{2}*\mu_{H_{2}}}{2*\mu_{H_{2}}}}$

$\omega_{D_{2}} = \frac{\omega_{H_{2}}}{\sqrt{2}} = \frac{1.29 \cdot 10^{14} Hz}{\sqrt{2}} = 9.12 \cdot 10^{13} Hz$

Therefore, the vibrational frequency of the deuterium molecule is 9.12x10¹³ Hz.

I hope it helps you!

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