Question

The limiting line in Balmer series will have a frequency of (Rydberg constant, $R_{\infty }=3.29\times 10^{15}$ cycles/s)

• A. $8.22\times 10^{14}{s}^{-1}$
• B. $3.29\times 10^{15}{s}^{-1}$
• C. $3.65\times 10^{14}{s}^{-1}$
• D. $5.26\times 10^{13}{s}^{-1}$

Answer 1

Answer: (A)

$\bar{v}=\frac{1}{\lambda }=R_{H}Z\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
In Balmer series $n_1=2\&n_2=3,4,\mathrm{5.....}$ Last line of the spectrum is called series limit
Limiting line is the line of shortest wavelength and high energy when $n_2=\infty$
$\therefore \bar{v}=\frac{1}{\lambda }=\frac{R_H}{n_1^2}=\frac{3.29\times 10^{15}}{2^2}=\frac{3.29\times 10^{15}}{4}$
$=8.22\times 10^{14}{s}^{-1}$