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Ionisation potential of hydrogen atom is $13.6 \mathrm{eV}$. Hydrogen atom in ground state is excited by monochromatic light of energy $12.1 \mathrm{eV}$. The spectral lines emitted by hydrogen according to Bohr's theory will be
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three
Since ionisation potential of hydrogen atom is $13.6 \mathrm{eV}$.
$\therefore \quad E_1=-13.6 \mathrm{eV}$
Now, $E_n-E_1=\frac{-13.6}{n^2}-(-13.6)=12.1$
$\begin{aligned} & \frac{-13.6}{n^2}+13.6=12.1 \\ & \therefore \quad n=3\end{aligned}$
After absorbing $12.1 \mathrm{eV}$, the electron of $\mathrm{H}$-atom is excited to $3^{\text {rd }}$ shell.
Thus, possible transitions are 3 i.e., $3 \rightarrow 2$, $2 \rightarrow 1$ and $3 \rightarrow 1$
$\therefore \quad E_1=-13.6 \mathrm{eV}$
Now, $E_n-E_1=\frac{-13.6}{n^2}-(-13.6)=12.1$
$\begin{aligned} & \frac{-13.6}{n^2}+13.6=12.1 \\ & \therefore \quad n=3\end{aligned}$
After absorbing $12.1 \mathrm{eV}$, the electron of $\mathrm{H}$-atom is excited to $3^{\text {rd }}$ shell.
Thus, possible transitions are 3 i.e., $3 \rightarrow 2$, $2 \rightarrow 1$ and $3 \rightarrow 1$
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