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## Homework Statement

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

## Homework Equations

EK = −(Z − 1)2(13.6 eV)

EL = −Zeff2E2

λ =

hc

ΔE

## 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