The Kalpha x-ray is emitted when an electron undergoes a transition from the L shell (n = 2) to the K shell (n = 1). Use the following equations to calculate the wavelength of the Kalpha x-ray from a iron target (Z = 26), where E0 is the ground-state energy.
The problem is the number im using in my (Z - #) for the E_L shell. I am missing the concept here of how to know which number to use. If Someone can explain that part to me, this question will be answered
EK = −(Z − 1)2(13.6 eV)
EL = −Zeff2E2
The Attempt at a Solution
E-k = -(26 - 1)^2(13.6 eV) = -8500 eV
E_l = -(26 - 4)^2(13.6 eV/4) = -1645.6 eV
λ = (4.14e-15 eVs)(3e8 m/s) / -1643.6 eV + 8500 eV = .181 nm