Question

Calculate the frequency of the limiting line of Brackett series in the hydrogen spectrum.

• A. $2.06\times {10}^{14}{s}^{-1}$
• B. $6.02\times {10}^{10}{s}^{-1}$
• C. $4.52\times {10}^{16}{s}^{-1}$
• D. $2.06\times {10}^{12}{s}^{-1}$

Answer 1

The correct option is A $2.06\times {10}^{14}{s}^{-1}$

For Brackett series in the Hydrogen spectrum,
$\frac{1}{\lambda }=R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
where $R_H=109677c{m}^{-1}$
For Brackett series $n_1=4$ and limiting line $n_2=\infty$
$\frac{1}{\lambda }=109677(\frac{1}{4^2}-\frac{1}{{\infty }^2})$
$\lambda =\frac{16}{109677}cm$ = $\frac{16}{109677\times 100}$m
Now, the relation between wavelength and frequency is given by c = $\nu \lambda$
where c = speed of light and $\nu$ is the frequency of the wavelength.
$3\times {10}^8=\nu \times \frac{16}{109677\times 100}$
On solving, frequency = $2.06\times {10}^{14}{s}^{-1}$