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(A) According to Bohr's energy formula,$\Delta E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) \ 	ext{eV/atom}$.
For the transition fr

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$32 / 27$

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(A) According to Bohr's energy formula,$\Delta E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) \ 	ext{eV/atom}$.
For the transition from $n=1$ to $n=3$:
$\Delta E_{1}$ ${ightarrow 3} = 13.6 \left[ \frac{1}{1^2} - \frac{1}{3^2} ight] = 13.6 \left[ 1 - \frac{1}{9} ight] = 13.6 \left[ \frac{8}{9} ight]$.
For the transition from $n=1$ to $n=2$:
$\Delta E_{1}$ ${ightarrow 2} = 13.6 \left[ \frac{1}{1^2} - \frac{1}{2^2} ight] = 13.6 \left[ 1 - \frac{1}{4} ight] = 13.6 \left[ \frac{3}{4} ight]$.
Therefore,the ratio is:
$\frac{\Delta E_{1 	o 3}}{\Delta E_{1 	o 2}} = \frac{13.6 \left[ \frac{8}{9} ight]}{13.6 \left[ \frac{3}{4} ight]} = \frac{8}{9} 	imes \frac{4}{3} = \frac{32}{27}$

The electron in a hydrogen atom is in the first Bohr orbit $(n=1)$. The ratio of transition energies,$E(n=1 \rightarrow n=3)$ to $E(n=1 \rightarrow n=2)$,is

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