Question

The K-shell ionization energy of iron is 8500 eV, and it's L-shell ionization energy is 2125 eV. What is the wavelength of Kalpha X-rays emmited by iron?

Answer 1

`1.95\cdot 10^(-10)m`

Step-by-step explanation:

First of all, we need to calculate the energy of the x-ray photon emitted during the transition from K-shell to L-shell, and this energy is equal to the difference in energy between the two levels:

`E=E_K-E_L=8500 eV-2125 eV=6375 eV`

Converting into Joules,

`E=6375 eV \cdot (1.6\cdot 10^(-19) J/eV)=1.02\cdot 10^(-15) J`

Now we know that the energy of the photon is related to its wavelength by:

`E=(hc)/(\lambda)`

where

h is the Planck constant

c is the speed of light

`\lambda` is the wavelength

Re-arranging the equation for
`\lambda`, we find

`\lambda=(hc)/(E)=((6.63\cdot 10^(-34)Js)(3\cdot 10^8 m/s))/(1.02\cdot 10^(-15) J)=1.95\cdot 10^(-10)m`