The wavelength of the Kalpha line in copper i | vedclass

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(B) The ionization energy $E$ required to remove a $K$-shell electron is given by the energy of the photon corresponding to the $K$-edge wavelength. U

Vedclass · Accepted Answer

$12.9 	imes 10^{-16}$

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(B) The ionization energy $E$ required to remove a $K$-shell electron is given by the energy of the photon corresponding to the $K$-edge wavelength. Using the relation $E = \frac{hc}{\lambda}$:Given: $h = 6.63 	imes 10^{-34} \ J \cdot s$,$c = 3 	imes 10^8 \ m/s$,and $\lambda = 1.54 \ \mathring{A} = 1.54 	imes 10^{-10} \ m$.$E = \frac{(6.63 	imes 10^{-34}) 	imes (3 	imes 10^8)}{1.54 	imes 10^{-10}}$$E = \frac{19.89 	imes 10^{-26}}{1.54 	imes 10^{-10}}$$E \approx 12.91 	imes 10^{-16} \ J$.

The wavelength of the $K\alpha$ line in copper is $1.54 \ \mathring{A}$. The ionization energy of the $K$ electron in copper in Joules is:

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