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(C) The energy of a photon with wavelength $\lambda = 800 \ nm$ is given by $E = \frac{hc}{\lambda} = \frac{1240}{800} = 1.55 \ eV$.Since only photons

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$3^{rd}$ excited state

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(C) The energy of a photon with wavelength $\lambda = 800 \ nm$ is given by $E = \frac{hc}{\lambda} = \frac{1240}{800} = 1.55 \ eV$.Since only photons with $\lambda > 800 \ nm$ pass through the filter,the energy of the emitted photons must be $E The energy levels of the hydrogen atom are $E_n = -\frac{13.6}{n^2} \ eV$.For $n=1, 2, 3, 4, 5$,the energies are $-13.6, -3.4, -1.51, -0.85, -0.54 \ eV$ respectively.The energy difference for the transition $n=4 	o n=3$ is $\Delta E = E_4 - E_3 = -0.85 - (-1.51) = 0.66 \ eV$.Since $0.66 \ eV The next possible transition $n=5 	o n=4$ has $\Delta E = -0.54 - (-0.85) = 0.31 \ eV$,which also passes. However,if the sample were in the $4^{th}$ excited state $(n=5)$,multiple transitions would result in photons passing through the filter. Given that only one type of photon passes,the sample must be in the $3^{rd}$ excited state $(n=4)$.

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