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In hydrogen spectrum, if the shortest wavelength in Balmer series is $\lambda$, the shortest wavelength in Brackett series is
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Verified Answer
The correct answer is:
$4 \lambda$
For Balmer series,
$\frac{1}{\lambda_{\text {Balmer }}}=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right)$, where $n=3,4,5 \ldots$
For shortest wavelength, $n=\infty$
$\begin{array}{rlrl}\Rightarrow & \frac{1}{\lambda} & =R\left(\frac{1}{2^2}\right)=\frac{R}{4} \\ \text { or } & \lambda=\frac{4}{R}\end{array}$
For Brackett series,
$\frac{1}{\lambda_{\text {Brackett }}}=R\left(\frac{1}{4^2}-\frac{1}{n^2}\right)$, where $n=5,6,7, \ldots$
Again, for shortest wavelength, $n=\infty$
$\begin{aligned} & \Rightarrow \frac{1}{\lambda_{\text {Brackett }}}=R\left(\frac{1}{4^2}\right)=\frac{R}{16} \\ & \text { or } \lambda_{\text {Brackett }}=\frac{16}{R}=4\left(\frac{4}{R}\right)=4 \lambda\end{aligned}$
[From Eq. (i)]
$\frac{1}{\lambda_{\text {Balmer }}}=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right)$, where $n=3,4,5 \ldots$
For shortest wavelength, $n=\infty$
$\begin{array}{rlrl}\Rightarrow & \frac{1}{\lambda} & =R\left(\frac{1}{2^2}\right)=\frac{R}{4} \\ \text { or } & \lambda=\frac{4}{R}\end{array}$
For Brackett series,
$\frac{1}{\lambda_{\text {Brackett }}}=R\left(\frac{1}{4^2}-\frac{1}{n^2}\right)$, where $n=5,6,7, \ldots$
Again, for shortest wavelength, $n=\infty$
$\begin{aligned} & \Rightarrow \frac{1}{\lambda_{\text {Brackett }}}=R\left(\frac{1}{4^2}\right)=\frac{R}{16} \\ & \text { or } \lambda_{\text {Brackett }}=\frac{16}{R}=4\left(\frac{4}{R}\right)=4 \lambda\end{aligned}$
[From Eq. (i)]
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