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

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The discussion focuses on calculating the diameter of a hydrogen atom with a principal quantum number n=732 and the wavenumber for the transition from n=732 to n=731. The energy transition was calculated as E_trans=6.95×10^-5 eV. The wavelength was determined to be λ=0.0178 m, leading to a wavenumber of ν=56.18 m^-1 using the formula ν=1/λ. The diameter of the hydrogen atom was found to be d_n=5.67×10^-5 m, confirming the calculations are consistent and relevant.
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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



E_{trans}= \frac{-E_{h}}{(732)^2}-\frac{-E_{h}}{(731)^2}}



The Attempt at a Solution



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

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

Homework Statement



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

Homework Statement



Is that a relevant result?







The Attempt at a Solution



the diameter will be=
d_{n}=2\cdot r_{0}\cdot n^2= 2\cdot (732)^2\cdot 5.291\cdot10^{-5} = 5.67\cdot10^{-5} m
 
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