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(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -13.6 / n^2 \, eV$.For $n=3 \rightarrow n=2$: $\Delta E = E_

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$n=3 \rightarrow n=2$

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(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -13.6 / n^2 \, eV$.For $n=3 \rightarrow n=2$: $\Delta E = E_3 - E_2 = -13.6(1/9 - 1/4) = -13.6(-5/36) \approx 1.89 \, eV$.For $n=4 \rightarrow n=3$: $\Delta E = E_4 - E_3 = -13.6(1/16 - 1/9) = -13.6(-7/144) \approx 0.66 \, eV$.For $n=5 \rightarrow n=4$: $\Delta E = E_5 - E_4 = -13.6(1/25 - 1/16) = -13.6(-9/400) \approx 0.31 \, eV$.Comparing the values,the transition $n=3 \rightarrow n=2$ releases the maximum energy.

Which transition in a $H$ atom emits the maximum energy?

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