Q156 / 160 · TS EAMCET 2019 (04 May Shift 2) · Physics
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
- Correct answerA6
- B4
- C5
- D7
Solution
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 $n_B$.
$\begin{aligned}<br />& \text { Hence, }(E)_{n_B}-(E)_{n_A}=\frac{|E|}{12} \\<br />& \frac{-|E|}{n_B^2}-\left\{\frac{-|E|}{n_A^2}\right\}=\frac{|E|}{12} \\<br />& \frac{-1}{n_B^2}+\frac{1}{n_A^2}=\frac{1}{12} \Rightarrow-\frac{1}{n_B^2}=\frac{1}{12}-\frac{1}{n_A^2} \\<br />& =\frac{1}{12}-\frac{1}{9} \Rightarrow-\frac{1}{n_B^2}=-\frac{1}{36} \\<br />& \Rightarrow n_B^2=36 \\<br />& \therefore n_B=6<br />\end{aligned}$
$\therefore$ 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 $n_B$.
$\begin{aligned}<br />& \text { Hence, }(E)_{n_B}-(E)_{n_A}=\frac{|E|}{12} \\<br />& \frac{-|E|}{n_B^2}-\left\{\frac{-|E|}{n_A^2}\right\}=\frac{|E|}{12} \\<br />& \frac{-1}{n_B^2}+\frac{1}{n_A^2}=\frac{1}{12} \Rightarrow-\frac{1}{n_B^2}=\frac{1}{12}-\frac{1}{n_A^2} \\<br />& =\frac{1}{12}-\frac{1}{9} \Rightarrow-\frac{1}{n_B^2}=-\frac{1}{36} \\<br />& \Rightarrow n_B^2=36 \\<br />& \therefore n_B=6<br />\end{aligned}$
Hence, the correct answer is option: 'A'.