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The ionization energy of the electron in the hydrogen atom in its ground state is $13.6 \mathrm{eV}$. The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength of emitted radiation corresponds to the transition between
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The correct answer is:
$n=4$ to $n=3$ states
Key Idea Number of spectral lines abtained due to transition of electron from $\boldsymbol{n}^{\text {th }}$ orbit to lower orbit is $\mathrm{N}=\frac{\mathrm{n}(\mathrm{n}-1)}{2}$ and for maximum wavelength the difference between the orbits of the series should be minimum.
Number of spectral lines $\mathrm{N}=\frac{\mathrm{n}(\mathrm{n}-1)}{2}$
$$
\begin{aligned}
\Rightarrow & \frac{n(n-1)}{2} & =6 \\
\text { or } & n^2-n-12 & =0 \\
\text { or } & (n-4)(n+3) & =0 \\
\text { or } & n & =4
\end{aligned}
$$
Now as the first line of the series has the maximum wavelength, therefore electron jumps from the $4^{\text {th }}$ orbit to the third orbit.
Number of spectral lines $\mathrm{N}=\frac{\mathrm{n}(\mathrm{n}-1)}{2}$
$$
\begin{aligned}
\Rightarrow & \frac{n(n-1)}{2} & =6 \\
\text { or } & n^2-n-12 & =0 \\
\text { or } & (n-4)(n+3) & =0 \\
\text { or } & n & =4
\end{aligned}
$$
Now as the first line of the series has the maximum wavelength, therefore electron jumps from the $4^{\text {th }}$ orbit to the third orbit.
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