Question

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.

Answer 1

Answer:

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!

Answer 2

The vibrational frequency of D₂ is :   9.12 * 10¹³ Hz

Given that:

Vibrational frequency ( w ) = $\sqrt{\frac{k}{u} }$

u = reduced mass

The reduced mass of a diatomic molecule is expressed as

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

Where : Ma and Mb are the atomic masses of mass A and mass B

First step : expressing the vibrational frequency of the hydrogen molecule

wH₂ = $\sqrt{\frac{k}{uH_{2} } }$   ----- ( i )

from the equation

k = ( wH₂ )² * uH₂ ---- ( ii )

Next step : expressing the vibrational frequency of the deuterium molecule.

wD₂ = $\sqrt{\frac{k}{uD_{2} } }$  ---- ( iii )

Insert equation ( ii ) into equation ( iii )

wD₂ = $\frac{wH_{2} }{\sqrt{2} }$  = ( 1.29 * 10¹⁴ ) / ( √2 ) = 9.12 * 10¹³ Hz

Hence we can conclude that The vibrational frequency of D₂ is :   9.12 * 10¹³ Hz.

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