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(D) The frequency of the emitted photon is given by the relation $\Delta E = h
u$,where $\Delta E$ is the energy difference between the two energy le

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$n=2$ to $n=1$

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(D) The frequency of the emitted photon is given by the relation $\Delta E = h
u$,where $\Delta E$ is the energy difference between the two energy levels.For a hydrogen atom,the energy of the $n^{th}$ level is $E_n = -\frac{13.6 \ eV}{n^2}$.The energy difference for a transition from $n_i$ to $n_f$ is $\Delta E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight) \ eV$.To have the highest frequency,we need the largest energy difference $\Delta E$.Calculating $\Delta E$ for each option:$A$: $n=1$ to $n=3$ (Absorption,not emission)$B$: $n=2$ to $n=4$ (Absorption,not emission)$C$: $n=5$ to $n=3$: $\Delta E = 13.6 \left( \frac{1}{3^2} - \frac{1}{5^2} ight) = 13.6 \left( \frac{1}{9} - \frac{1}{25} ight) \approx 13.6 	imes 0.071 \approx 0.96 \ eV$.$D$: $n=2$ to $n=1$: $\Delta E = 13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = 13.6 \left( 1 - 0.25 ight) = 13.6 	imes 0.75 = 10.2 \ eV$.Since $10.2 \ eV > 0.96 \ eV$,the transition $n=2$ to $n=1$ releases the most energy and thus emits the photon with the highest frequency.

Which of the following transitions in a hydrogen atom emits photons of the highest frequency? ($n=$ principal quantum number)

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