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If the electron in hydrogen atom jumps from third orbit to second orbit, the wavelength of the emitted radiation is given by:
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Verified Answer
The correct answer is:
\(\lambda=\frac{36}{5 R}\)
We have, \(\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)\)
\(\begin{aligned}
& \frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\
& \frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{9}\right) \\
& \frac{1}{\lambda}=\mathrm{R}\left(\frac{9-4}{36}\right) \\
& \frac{1}{\lambda}=\mathrm{R} \times \frac{5}{36} \\
& \lambda=\frac{36}{5 \mathrm{R}}
\end{aligned}\)
\(\begin{aligned}
& \frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\
& \frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{9}\right) \\
& \frac{1}{\lambda}=\mathrm{R}\left(\frac{9-4}{36}\right) \\
& \frac{1}{\lambda}=\mathrm{R} \times \frac{5}{36} \\
& \lambda=\frac{36}{5 \mathrm{R}}
\end{aligned}\)
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