given ion like hydrogen that it electron are in the Excited state n.
the ionization energy of this ion from the Excited state n is 7.65eV.

irradiation in light with frequency of 6.65x10^14 sec^-1 making to jump to Excited state n+1.

how I can calculate what is the Excited state n and who is the ion of this possibility:He+ , Li2+


tell me if I correct here: the E=Hf is 6.65x10^14 sec^-1

thanks for help. And sorry for my bad english :)

The ionisation energy for the ion is 7.65 eV.

The ionisation energy for the next excited state (n+1) is given by E = hf.

In your answer you the answer as the frequency? :confused:

I get E = 3.99 x 10^-13 J = 2.49 eV

I don't know if there is a relationship (formula relating) between n and the energy, but if not, there should be a table when you have some value of:

n = 7.65 eV
n + 1 = 2.49 eV

And from there, deduce which of He^+ or Li^2+ is correct.

I will try to ask in a little better english.

Given a ion Hydrogen-like that it's electron is in an excited state n.
The ionization energy of the ion from the excited state n is 7.65eV.
Radiating the electron in light frequency of 6.65x10^14sec-1 transfers it to excited state n+1.

I need to calculate what is the excited state n, and which is ion: He+ or Li2+ ?

If we know that radiating the electron in light frequency of 6.65x10^14sec-1 transfers it to excited state n+1.
Does it not means that 6.65x10^14sec-1 = E=hf?

I need to use the Bohr model.

Thank you very much!

6.65x10^14sec-1 = E = hf

Energy is not frequency...

yes that correct.
by the way did you used with the formula -13.6*Z^2/n^2 ?
if not can you give me the formula that you used to calculate it ?
thank you.

Wait... I did a mistake earlier >.< I think there was some number on my calculator from a previous work which got in the calculation.

I used the simple E = hf

E = hf = 6.63\times10^{-34}\ \times\ 6.65\times10^{14}\ =\ 4.41 \times 10^{-19}\ J

Now, using:

1\ J\ =\ 1.6\times 10^{-19}\ eV

4.41\times10^{-19}\ J\ =\ 2.75 \ eV

Meaning that the ion gained 2.75 eV to get on to the n+1 state.

The ionisation energy from the n+1 state is thus = 7.65 - 2.75 = 4.90 eV

Okay, so the ionisation energy for:

n = 7.65 eV
n + 1 = 4.90 eV

Now, for this formula you gave, this is the first time I see it and I did some research.

Okay, so let's try He^+

Z = 2

E_n = \frac{-13.6 \times 2^2}{n^2}

E_{n+1} = \frac{-13.6 \times 2^2}{(n+1)^2}

If En and En+1 give satisfactory values of n, then He^+ is the one.

If not, we try with Li^+

Z = 3

So many thanks :)
Again help me a lot