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

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Deuterium's vibrational frequency can be calculated using the formula w = √(k/m), where k is the spring constant and m is the mass of the molecule. Given that deuterium's mass is twice that of hydrogen, its vibrational frequency will be lower than that of helium, which is 1.34x10^14 Hz. The relationship between the frequencies can be expressed as f1/f2 = √(m2/m1), indicating that deuterium's frequency is inversely proportional to the square root of its mass. Therefore, the vibrational frequency of deuterium is approximately 9.48x10^13 Hz. This calculation assumes the spring constant remains constant for both species.
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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: \omega = 2\pi f = \sqrt{\frac{k}{m}

So what is f_{1}/f_{2}?

AM
 
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