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 :

skyluke89

Answer:

[tex]1.95\cdot 10^{-10}m[/tex]

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:

[tex]E=E_K-E_L=8500 eV-2125 eV=6375 eV[/tex]

Converting into Joules,

[tex]E=6375 eV \cdot (1.6\cdot 10^{-19} J/eV)=1.02\cdot 10^{-15} J[/tex]

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

[tex]E=\frac{hc}{\lambda}[/tex]

where

h is the Planck constant

c is the speed of light

[tex]\lambda[/tex] is the wavelength

Re-arranging the equation for [tex]\lambda[/tex], we find

[tex]\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.02\cdot 10^{-15} J}=1.95\cdot 10^{-10}m[/tex]

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