Question

The limiting line in Balmer series will have a frequency of (Rydberg constant, $R_{\infty}=3.29\times10^{15}$ 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

The Correct Option is A

$\bar{v}=\frac{1}{\lambda}=R_{H}Z\left(\frac{1}{n^{2}_{1}}-\frac{1}{n^{2}_{2}}\right)$ In Balmer series $n_{1} = 2 \& n_{2}=3, 4, 5 .....$ Last line of the spectrum is called series limit Limiting line is the line of shortest wavelength and high energy when $n_{2}=\infty$ $\therefore \bar{v}=\frac{1}{\lambda}=\frac{R_{H}}{n^{2}_{1}}=\frac{3.29\times10^{15}}{2^{2}}=\frac{3.29\times10^{15}}{4}$ $=8.22\times10^{14}s^{-1}$