Question

In hydrogen spectrum if the shortest wavelength in Balmer series is $\lambda$, the shortest wavelength in Bracket series is

• A. $\lambda$
• B. $\frac{\lambda }{2}$
• C. $4\lambda$
• D. $9\lambda$

Answer 1

Answer: (C)

For Balmer series,
$\frac{1}{{\lambda }_{\text{Balmer }}}=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right)$,
where $n=3,4,5\dots$
For shortest wavelength, $n=\infty$
$\Rightarrow \frac{1}{\lambda }=R\left(\frac{1}{2^2}\right)=\frac{R}{4}$
or $\lambda =\frac{4}{R}\dots .(i)$
For Brackett series,
$\frac{1}{{\lambda }_{\text{Brackett }}}=R\left(\frac{1}{4^2}-\frac{1}{n^2}\right)$,
where $n=5,6,7,\dots$
Again, for shortest wavelength, $n=\infty$
$\Rightarrow \frac{1}{{\lambda }_{\text{Brackett }}}=R\left(\frac{1}{4^2}\right)=\frac{R}{16}$
or ${\lambda }_{\text{Brackett }}=\frac{16}{R}=4\left(\frac{4}{R}\right)=4\lambda [$ From Eq.(i)]