[SumerianKing]

I have done the question but it seems wrong.

Q- Using the Rydberg equation calculate the ionisation energy of an electron in the 4s orbital of hydrogen.

Ok first of all since its hydrogen I should use Rh right? So Rh= 2.18x10^-18

Now since its asking for ionization energy the first n1 value is 1
And the second n2 value im saying is 4, as its in the 4s orbital. Im not sure??

So putting this together is 2.18x10^-18 X (1/1^2 – ¼^2)

So...  2.18x10^-18 X 0.1875 = 4.08x10^-19

Now i got the frequency, Now to work out Ionization energy I obviously use E=hv where h=6.626x10^-34

So now 4.08x10^-19 X 6.626x10^-34 = 2.70x10^-52 !!!!

Is that correct? The number is incredibly small? And I don’t know whether this is the correct way of working it out for the 4s orbital , or weather all the values I have inputted are correct. I would appreciate it so much for some help 

[Schrödinger]

So putting this together is 2.18x10^-18 X (1/1^2 – ¼^2)

So...  2.18x10^-18 X 0.1875 = 4.08x10^-19

Check those figures again.

The answer you get from the Rydberg equation is ionization energy, not frequency.

[SumerianKing]

So putting this together is 2.18x10^-18 X (1/1^2 – ¼^2)

So...  2.18x10^-18 X 0.1875 = 4.08x10^-19

Check those figures again.

The answer you get from the Rydberg equation is ionization energy, not frequency.

Oh i see where i went wrong. Ok i calculated the new rydberg equation, its 2.04x10^-18

How is the answer for ionization energy Rydberg equation gives me the frequency. In my notes and reading chem books it tells me "after working out frequency from rydberg equation, then work out ionization energy by using E=hv" ?? Thanks allot

[Schrödinger]

Some times the Rydberg equation is used in another form, with the LHS being wavenumber (reciprocal of wavelength) and the RHS containing an additional 'hc' factor in the denominator (h = Planck's constant and c=speed of light). Although they're both the one and the same, it is possible that your book was referring to this form.