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(A) The energy of a photon is given by $\Delta E = \frac{hc}{\lambda}$.Rearranging for wavelength,we get $\lambda = \frac{hc}{\Delta E}$.Given $\Delta

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$\frac{12395}{\Delta E} 	imes 10^{-10} \ m$

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(A) The energy of a photon is given by $\Delta E = \frac{hc}{\lambda}$.Rearranging for wavelength,we get $\lambda = \frac{hc}{\Delta E}$.Given $\Delta E$ is in electron volts $(eV)$,we use $h = 6.626 	imes 10^{-34} \ J \ s$,$c = 3 	imes 10^{8} \ m/s$,and $1 \ eV = 1.602 	imes 10^{-19} \ J$.$\lambda = \frac{6.626 	imes 10^{-34} 	imes 3 	imes 10^{8}}{\Delta E 	imes 1.602 	imes 10^{-19}} \ m$.$\lambda \approx \frac{12395}{\Delta E} 	imes 10^{-10} \ m$.

When electronic transition occurs from a higher energy state to a lower energy state with an energy difference equal to $\Delta E$ electron volts,the wavelength of the emitted line is approximately equal to

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