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In hydrogen spectrum, the shortest wavelength in the Balmer series is $\lambda$. The shortest wavelength in the Bracket series is :
A
$16\lambda$
B
$2\lambda$
C
$4\lambda$
D
$9\lambda$
Explanation
Series limit gives $\lambda=\frac{4}{R}$ (Balmer) and $\lambda'=\frac{16}{R}$ (Brackett), so $\lambda'=4\lambda$.
Detailed Solution
$\because\frac{1}{\lambda}=R\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)$
Balmer: $\frac{1}{\lambda}=R\left(\frac{1}{2^2}\right)\Rightarrow\lambda=\frac{4}{R}$
Brackett: $\frac{1}{\lambda'}=R\left(\frac{1}{4^2}\right)\Rightarrow\lambda'=\frac{16}{R}$
$\Rightarrow\lambda'=4\lambda$