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

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(C) The energy of a photon is given by $E = \frac{hc}{\lambda}$.To express $E$ in $eV$ and $\lambda$ in $\mathring{A}$,we use:$E(eV) = \frac{(6.626

Vedclass · Accepted Answer

$E = 12.4/\lambda$

Vedclass · Answer

(C) The energy of a photon is given by $E = \frac{hc}{\lambda}$.To express $E$ in $eV$ and $\lambda$ in $\mathring{A}$,we use:$E(eV) = \frac{(6.626 	imes 10^{-34} 	ext{ J s}) 	imes (3 	imes 10^8 	ext{ m/s})}{(1.602 	imes 10^{-19} 	ext{ J/eV}) 	imes (\lambda 	imes 10^{-10} 	ext{ m})}$$E(eV) \approx \frac{12400}{\lambda(\mathring{A})}$.To convert this to $KeV$,we divide by $1000$:$E(KeV) = \frac{12400}{1000 	imes \lambda(\mathring{A})} = \frac{12.4}{\lambda(\mathring{A})}$.Thus,the correct relation is $E = 12.4/\lambda$.

If we express the energy of a photon in $KeV$ and the wavelength in $\mathring{A}$,then the energy of a photon can be calculated from the relation:

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