How Does Doubling Molecular Mass Affect Vibration Frequency?

AI Thread Summary
Doubling the molecular mass from hydrogen (H2) to deuterium (D2) significantly impacts the vibrational frequency. The vibrational frequency of H2 is given as 1.35x10^14 Hz, and using the same spring constant for both molecules, the frequency for D2 is calculated. The calculations show that the frequency for D2 is approximately 0.1125 Hz, indicating a substantial decrease due to the increased mass. This relationship highlights the inverse square root dependence of frequency on mass in molecular vibrations. The discussion emphasizes the importance of understanding molecular mass in vibrational frequency calculations.
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1. The mass of the deuterium molecule D2 is twice the mass of Hydrogen molecule H2. If the vibrational frequency of H2 is 1.35x10^14 Hz, what is the vibrational frequency of D2, assuming the "spring constant" of attracting forces is the same for two species? Answer in unit of Hz.



Homework Equations



f=(w)/(2pie)
w=sq root (k/m)
k=m(w^2)
w is phrased as omega

The Attempt at a Solution



H2 :
f=1.35x10^14
(f)(2pie)=w = 8.48x10^14
k=m(w^2)=7.19x10^29(m)

D2 :
f=? <--solve
m=2k=2(7.19x10^29)m=1.439x10^30(m)
k=7.19m
w=sq root (k/m) = sq root [( 7.19x10^29m)/1.439x10^30m)]=w=0.707
f = 0.707/2pie = 0.1125 Hz
check?
 
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w=sq root (k/m) = sq root [( 7.19x10^29m)/1.439x10^30m)]=w=0.707

The denominator should be 2m, since the deuterium is twice as massive as the hydrogen.
 
H2:
k=m(w^2)=7.19x10^29(m)
D2:
m=2k=2(7.19x10^29)m=1.439x10^30(m)
 
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