The wavelength of the {K_alpha } line in copp | vedclass

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Question

(B) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.Here,$h = 6.6 	imes 10^{-34} \ J \cdot s$ (Planck's constant),$c = 3 	im

Vedclass · Accepted Answer

$12.9 	imes 10^{-16}$

Vedclass · Answer

(B) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.Here,$h = 6.6 	imes 10^{-34} \ J \cdot s$ (Planck's constant),$c = 3 	imes 10^8 \ m/s$ (speed of light),and $\lambda = 1.54 \mathring A = 1.54 	imes 10^{-10} \ m$.Substituting these values into the formula:$E = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{1.54 	imes 10^{-10}} \ J$$E = \frac{19.8 	imes 10^{-26}}{1.54 	imes 10^{-10}} \ J$$E \approx 12.857 	imes 10^{-16} \ J \approx 12.9 	imes 10^{-16} \ J$.

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

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