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(C) The energy of a photon is given by $E = \frac{hc}{\lambda}$ (in Joules). To convert this to $eV$,we divide by the elementary charge $e$: $E_{(eV)}

Vedclass · Accepted Answer

$E = 12.4/\lambda$

Vedclass · Answer

(C) The energy of a photon is given by $E = \frac{hc}{\lambda}$ (in Joules). To convert this to $eV$,we divide by the elementary charge $e$: $E_{(eV)} = \frac{hc}{e\lambda}$. Substituting the values $h = 6.626 	imes 10^{-34} \, J\cdot s$,$c = 3 	imes 10^8 \, m/s$,and $e = 1.6 	imes 10^{-19} \, C$,and expressing $\lambda$ in $\mathring{A}$ $(1 \, \mathring{A} = 10^{-10} \, m)$: $E_{(eV)} = \frac{6.626 	imes 10^{-34} 	imes 3 	imes 10^8}{1.6 	imes 10^{-19} 	imes \lambda_{(\mathring{A})} 	imes 10^{-10}} \approx \frac{12400}{\lambda_{(\mathring{A})}} \, eV$. Since $1 \, KeV = 1000 \, eV$,we have $E_{(KeV)} = \frac{12400}{1000 	imes \lambda_{(\mathring{A})}} = \frac{12.4}{\lambda_{(\mathring{A})}}$. Thus,the energy in $KeV$ is given by $E = \frac{12.4}{\lambda}$.

If the energy of a photon is expressed in units of $KeV$ and the wavelength in units of $\mathring{A}$,then the energy of the photon can be calculated by:

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