# Easy - determining n from orbital radius

## Homework Statement

For the hydrogen atom, determine the principle quantum number corresponding to an orbital of radius Rn=0.01mm

## Homework Equations

I'm using the following formula:

Rn=(n^2)(Ao)/Z

Where Rn is the orbital radius (0.01mm),
n is the principle quantum number (wanted)
Ao is a constant (0.5*10^-10)
and Z is the atomic number (for hydrogen =1)

## The Attempt at a Solution

By converting Rn into metres and subbing in all the other values I obtain a value of n=447.2135955.

Obviously there are two discreptancies:
1) n is not a whole number
2) it is abnormally large

Can anybody show me where I have gone wrong?

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Dick
Homework Helper
i) it's not a whole number because 0.01mm is an approximate value. ii) It's abnormally large because an 0.01mm hydrogen atom is abnormally large. It doesn't seem out of line to me.

Would I answer this with a rounded value then?

A lead on question asks: for this value of principle quantum number, find the correspongind excited state energy.

Thus E=-13.6Z^2/(n^2) becomes

E=-13.6/(447.2135955^2)=-6.8*10^-5 eV

(notice for this part I did not use the rounded value)

Dick
Homework Helper
There is no orbital at exactly 0.01mm. So yes, you round. It shouldn't matter much.

I have another question, slightly different topic

-Calculate the first excited state energies of He+, Li2+ and Be3+ compare them with the ground state energy of hydrogen.

ground state means n=1, but how do i find n for the first excited state? does it differ for each of the 3 cases above?

Last edited:
Dick
Homework Helper
Yes, n=1 in all those cases. Look at your formula. What could change?

atomic number is going to be different for each of those cases. But since I want to find the first excited state of He+.... don't I need to use the relevant principle quantum number?

Not sure if I've missed something here, but how do I determine the principle quantum number (n) for the first excited state in the three above cases?

Dick