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(A) The energy of a photon absorbed or emitted is given by $\Delta E = h
u$,where $
u$ is the frequency. To absorb a photon of maximum frequency,the

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From $n = 1$ to $n = 4$

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(A) The energy of a photon absorbed or emitted is given by $\Delta E = h
u$,where $
u$ is the frequency. To absorb a photon of maximum frequency,the transition must involve the largest energy gap $(\Delta E)$. For a hydrogen atom,the energy of an orbit is $E_n = -13.6 / n^2 \ eV$. $(A)$ $n = 1$ to $n = 4$: $\Delta E = 13.6(1 - 1/16) = 13.6 	imes (15/16) = 12.75 \ eV$. $(B)$ $n = 2$ to $n = 1$: This is an emission transition (energy released,not absorbed). $(C)$ $n = 2$ to $n = 3$: $\Delta E = 13.6(1/4 - 1/9) = 13.6 	imes (5/36) \approx 1.89 \ eV$. $(D)$ $n = 3$ to $n = 2$: This is an emission transition. Comparing the absorption transitions,the transition from $n = 1$ to $n = 4$ involves the largest energy change,thus the maximum frequency photon is absorbed.

Photon of the maximum frequency will be absorbed in the transition (for $H$ atom):

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