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Ionisation potential of hydrogen atom is $13.6 \mathrm{eV}$. When hydrogen atoms in ground state are excited by a supply of $12.1 \mathrm{eV}$, then the number of spectral lines emitted by hydrogen atoms according to Bohr's theory is
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The correct answer is:
$3$
Given that, ionisation potential of hydrogen atom, $E_1=-13.6 \mathrm{eV}$
Energy supplied = Energy absorbed by hydrogen atom.
$\Delta E=12.1 \mathrm{eV}$
i.e. final energy of excited state,
$E_2=\frac{-13.6 \mathrm{eV}}{n^2}$
We know that, $E_2-E_1=\Delta E$
$\Rightarrow \quad E_2=\Delta E+E_1$
Substituting the value, we get
$\begin{aligned} & -\frac{13.6}{n^2}=[12.1+(-13.6)] \\ & -\frac{13.6}{n^2}=-1.50 \Rightarrow x^2=9.07 \Rightarrow n \simeq 3\end{aligned}$
Hence, by absorbing $12.1 \mathrm{eV}$ energy electron jump into 2nd excited state $(n=3)$.
Then, number of spectral lines emitted by $\mathrm{H}$-atom
$=\frac{n(n-1)}{2}=\frac{3(3-1)}{2}=3$
Energy supplied = Energy absorbed by hydrogen atom.
$\Delta E=12.1 \mathrm{eV}$
i.e. final energy of excited state,
$E_2=\frac{-13.6 \mathrm{eV}}{n^2}$
We know that, $E_2-E_1=\Delta E$
$\Rightarrow \quad E_2=\Delta E+E_1$
Substituting the value, we get
$\begin{aligned} & -\frac{13.6}{n^2}=[12.1+(-13.6)] \\ & -\frac{13.6}{n^2}=-1.50 \Rightarrow x^2=9.07 \Rightarrow n \simeq 3\end{aligned}$
Hence, by absorbing $12.1 \mathrm{eV}$ energy electron jump into 2nd excited state $(n=3)$.
Then, number of spectral lines emitted by $\mathrm{H}$-atom
$=\frac{n(n-1)}{2}=\frac{3(3-1)}{2}=3$
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