What Is the Radius of the n = 6 Bohr Orbit in O7+?

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To calculate the radius of the n = 6 Bohr orbit in O7+, the relevant formula must account for the nuclear charge, which is Z=8 for oxygen. The standard equation r(n) = (n^2) * a(b) applies to hydrogen-like atoms but needs modification for different atomic numbers. The correct radius for the n = 6 orbit in O7+ is 238 pm, which requires using the adjusted formula that incorporates the atomic number. Users are encouraged to consult resources on "hydrogen-like" ions for a clearer understanding. Accurate calculations depend on recognizing the influence of nuclear charge on the orbit radius.
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Homework Statement


Calculate the radius of the n = 6 Bohr orbit in O7+(oxygen with 7 of its 8 electrons removed).
A) 190 pm B) 167 pm C) 238 pm D) 214 pm


Homework Equations



I believe that the relevant equation is
r(sub n)=(n^2)*a(sub b) where a(sub b)= Bohr's radius= 5.29*10^-11 m

The Attempt at a Solution


I calculated it for what I believed to be n=6, but I got an incorrect answer. The correct answer is (C) 238 pm, but I do not understand why.


Any help is greatly appreciated!
 
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Your formula is not quite complete. You need to take into account the charge in the nucleus (so the number of protons).
 
Welcome to PF :smile:

Your equation is for hydrogen, with a charge of +1 on the nucleus.
Since oxygen has a charge of Z=8 for the nucleus, that equation should be different, containing Z somehow.

Does your textbook discuss "hydrogen-like" or "hydrogenic" ions?

EDIT: ah, I should know better than to wait a 1/2 hour and then respond without refreshing the page. :redface:
 
Unfortunately,
I cannot find an equation that takes into account (Z). Anyone have the equation handy?
Thanks for all your help!
 
Thank you very much!
 

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