How Does Heavy Hydrogen Affect the Wavelength of the Balmer Series?

AI Thread Summary
The discussion focuses on calculating the difference in wavelength of the first line of the Balmer series between ordinary hydrogen and heavy hydrogen. The known wavelength for ordinary hydrogen is 656.1 nm, while the expected difference for heavy hydrogen is only 0.178 nm. The challenge arises from the fact that the reduced mass of the electron-nucleus system changes slightly, affecting the energy levels and thus the wavelengths. The Bohr model is suggested as a useful framework for understanding this deviation, emphasizing the role of reduced mass in the calculations. Ultimately, the participant resolves the issue with the calculations.
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Ok, for this one problem, I'm supposed to figure out the difference in wavelength between the first line of the Balmer series in ordinary hydrogen (M= 1.01 u) and in "heavy" hydrogen (M = 2.01 u). The balmer series is when the lowest n is 2, by the way. I already know that the wavelength for the first line of the series in ordinary hydrogen is 656.1 nm, but the answer to this question is supposed to be 0.178 nm, and I can't figure out any way to calculate a wavelength for heavy hydrogen that would give a difference that small. It seems like, no matter what I do, I keep getting half the wavelength of regular hydrogen. Anyone know what to do on this one?
 
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The change in wavelength would be very small because a "heavy hydrogen" nucleus has the same charge as a normal hydrogen nucleus, so it exerts roughly the same force on the electron. However, the center of mass of the electron-nucleus system will change. Can you see how this might effect the energy levels (and thus, the wavelengths)?
 
Check out the Bohr model and you will see that the electron mass plays a role in the energy level.

But the mass to use is the reduced mass mer=me.mp/(me+mp).
As you can see the change in mer is very small as mp/(me+mp) is nearly 1 as for both cases mp=1.01 or 2.01 as me = 1/1800 mp.

Using the Bohr model you will get a good indication of the magnitude of the deviation
 
Thanks guys! I got it.
 
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