What is the Wavenumber and Transition for an H-Atom with n=732 to n=731?

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SUMMARY

The discussion focuses on calculating the diameter of a hydrogen atom with a principal quantum number of n=732 and determining the wavenumber for the transition from n=732 to n=731. The energy transition was calculated using the formula E_{trans}= \frac{-E_{h}}{(732)^2}-\frac{-E_{h}}{(731)^2}, resulting in an energy value of 6.95 x 10^{-5} eV. The wavelength was subsequently calculated as 0.0178 m, leading to a wavenumber of 56.18 m^{-1} using the formula \nu=\frac{1}{\lambda}. The diameter of the hydrogen atom was found to be 5.67 x 10^{-5} m.

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Homework Statement


Calculate a diameter of an H-atom with n=732. Calculate also the value of the wavenumber corresponding to the transition from n=732 to n=731




Homework Equations



[tex]E_{trans}= \frac{-E_{h}}{(732)^2}-\frac{-E_{h}}{(731)^2}}[/tex]



The Attempt at a Solution



[tex]E_{trans}= \frac{-13.6}{(732)^2}-\frac{-13.6}{(731)^2}}=6.95\cdot10^{-5} eV[/tex]

[tex]E=\frac{hc}{\lambda}\Rightarrow \lambda=\frac{hc}{E}=\frac{4.135\cdot10^{-15}\cdot 3\cdot10^8}{6.95\cdot10^{-5}}=0.0178 m[/tex]

Homework Statement



now going back to the wavenumber, i was not sure which formula i should use...
[tex]k=\frac{2\pi}{\lambda} or \nu=\frac{1}{\lambda}[/tex]
I have chosen the second one and i got
[tex]\nu=\frac{1}{0.0178}=56.18 m^{-1}[/tex]

Homework Statement



Is that a relevant result?







The Attempt at a Solution



the diameter will be=
[tex]d_{n}=2\cdot r_{0}\cdot n^2= 2\cdot (732)^2\cdot 5.291\cdot10^{-5} = 5.67\cdot10^{-5} m[/tex]
 
Last edited:
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Seems OK :)
 

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