How Does Deuterium's Vibrational Frequency Compare to Helium's?

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SUMMARY

The vibrational frequency of deuterium can be calculated using the formula for vibrational frequency, \( \omega = \sqrt{\frac{k}{m}} \), where \( k \) is the spring constant and \( m \) is the mass of the molecule. Given that the mass of deuterium is twice that of hydrogen and the vibrational frequency of helium is 1.34x1014 Hz, the frequency of deuterium can be derived by comparing the two frequencies. The relationship \( f_{1}/f_{2} \) can be established to find the exact frequency of deuterium.

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The mass of the deuterium molecule is twice that of the hydrogen molecule. If the vibrational frequency of helium is 1.34x10^14 Hz, what is the vibrational frequncy of deuterium, assyming that the spring constant of attracting forces is the same for the two species? Answer in units of Hz.

frequency: w=square root of k/m

...?

Please help thank you.
 
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jdog6 said:
The mass of the deuterium molecule is twice that of the hydrogen molecule. If the vibrational frequency of helium is 1.34x10^14 Hz, what is the vibrational frequncy of deuterium, assyming that the spring constant of attracting forces is the same for the two species? Answer in units of Hz.
frequency: w=square root of k/m
You have correctly used: [tex]\omega = 2\pi f = \sqrt{\frac{k}{m}[/tex]

So what is [tex]f_{1}/f_{2}[/tex]?

AM
 

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