If we express the energy of a photon in KeV a | vedclass

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Question

(c) Energy of photon $E = \frac{{hc}}{\lambda }$ (Joules) $ = \frac{{hc}}{{e\lambda }}(eV)$
$ \Rightarrow \,\mathop E\limits_{(eV)} = \frac{{6.6 	im

Vedclass · Accepted Answer

$E = 12.4\,/\lambda $

Vedclass · Answer

(c) Energy of photon $E = \frac{{hc}}{\lambda }$ (Joules) $ = \frac{{hc}}{{e\lambda }}(eV)$
$ \Rightarrow \,\mathop E\limits_{(eV)} = \frac{{6.6 	imes {{10}^{ - 34}} 	imes 3 	imes {{10}^8}}}{{1.6 	imes {{10}^{ - 19}} 	imes \lambda ({ \mathring A})}} = \frac{{12375}}{{\lambda ({ \mathring A})}}$
$ \Rightarrow \,\,E(keV) = \frac{{12.37}}{{\lambda ({ \mathring A})}} \approx \frac{{12.4}}{\lambda }$

If we express the energy of a photon in $KeV$ and the wavelength in angstroms, then energy of a photon can be calculated from the relation

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