The energy of an X-ray photon of wavelength 1 | vedclass

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(C) The energy $E$ of a photon is given by the formula $E = \frac{hc}{\lambda}$.Given:$h = 6.6 	imes 10^{-34} \ J \cdot s$$c = 3 	imes 10^8 \ m/s$$\

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(C) The energy $E$ of a photon is given by the formula $E = \frac{hc}{\lambda}$.Given:$h = 6.6 	imes 10^{-34} \ J \cdot s$$c = 3 	imes 10^8 \ m/s$$\lambda = 1.65 \mathring{A} = 1.65 	imes 10^{-10} \ m$Substituting the values:$E = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{1.65 	imes 10^{-10}} \ J$$E = \frac{19.8 	imes 10^{-26}}{1.65 	imes 10^{-10}} \ J = 12 	imes 10^{-16} \ J$To convert this energy into $eV$,divide by $1.6 	imes 10^{-19} \ J/eV$:$E(eV) = \frac{12 	imes 10^{-16}}{1.6 	imes 10^{-19}} = 7.5 	imes 10^3 \ eV = 7.5 \ keV$.Alternatively,using the shortcut $E(eV) = \frac{12400}{\lambda(\mathring{A})}$,we get $E = \frac{12400}{1.65} \approx 7515 \ eV \approx 7.5 \ keV$.

The energy of an $X$-ray photon of wavelength $1.65 \mathring{A}$ is ....... $keV$. $(h = 6.6 \times 10^{-34} \ J \cdot s, \ c = 3 \times 10^8 \ m/s, \ 1 \ eV = 1.6 \times 10^{-19} \ J)$ (in $.5$)

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