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A hydrogen sample is prepared in a particular excited state $A$ of quantum number, $n_A=3$. The ground state energy of hydrogen atom is $-|E|$. The photons of energy $\frac{|E|}{12}$ are absorbed in the sample which results in the excitation of some electrons to excited state $B$ of quantum number $n_B$, whose value is
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6
Given, energy of hydrogen atom in ground state $=-|E|$ and $n_A=3$
$\therefore$ Energy of electrons in excited state, $n_A$
$(E)_{n_A}=\frac{-|E|}{n_A^2}$
Energy of electrons in excited state, $n_B$
$=(E)_{n B}=\frac{-(E)}{n_B^2}$
When sample absorbs the photon of energy, $\frac{|E|}{12}$, then its electrons reaches from energy state $n_A$ to energy state $n_B$.
$\begin{aligned}
& \text { Hence, }(E)_{n_B}-(E)_{n_A}=\frac{|E|}{12} \\
& \frac{-|E|}{n_B^2}-\left\{\frac{-|E|}{n_A^2}\right\}=\frac{|E|}{12} \\
& \frac{-1}{n_B^2}+\frac{1}{n_A^2}=\frac{1}{12} \Rightarrow-\frac{1}{n_B^2}=\frac{1}{12}-\frac{1}{n_A^2} \\
& =\frac{1}{12}-\frac{1}{9} \Rightarrow-\frac{1}{n_B^2}=-\frac{1}{36} \\
& \Rightarrow n_B^2=36 \\
& \therefore n_B=6
\end{aligned}$
$\therefore$ Energy of electrons in excited state, $n_A$
$(E)_{n_A}=\frac{-|E|}{n_A^2}$
Energy of electrons in excited state, $n_B$
$=(E)_{n B}=\frac{-(E)}{n_B^2}$
When sample absorbs the photon of energy, $\frac{|E|}{12}$, then its electrons reaches from energy state $n_A$ to energy state $n_B$.
$\begin{aligned}
& \text { Hence, }(E)_{n_B}-(E)_{n_A}=\frac{|E|}{12} \\
& \frac{-|E|}{n_B^2}-\left\{\frac{-|E|}{n_A^2}\right\}=\frac{|E|}{12} \\
& \frac{-1}{n_B^2}+\frac{1}{n_A^2}=\frac{1}{12} \Rightarrow-\frac{1}{n_B^2}=\frac{1}{12}-\frac{1}{n_A^2} \\
& =\frac{1}{12}-\frac{1}{9} \Rightarrow-\frac{1}{n_B^2}=-\frac{1}{36} \\
& \Rightarrow n_B^2=36 \\
& \therefore n_B=6
\end{aligned}$
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