Question

For any given series of spectral lines of atomic hydrogen, let $\Delta V=V_{max}-V_{min}$ be the difference in maximum and minimum frequencies in cm${}^{-1}$. The ratio $\Delta \overline{V_{\text{Lyman}}}/\Delta \overline{V_{\text{Balmer}}}$ is :

• A. 27 : 5
• B. 4 : 1
• C. 5 : 4
• D. 9 : 4

Answer 1

Answer: (D)

For Lyman
$\overline{V_{max}}{R}_H(\frac{1}{1^2}-\frac{1}{{\infty }^2})=R_H$
$\overline{V_{max}}{R}_H(\frac{1}{1^2}-\frac{1}{2^2})=\frac{3}{4}{R}_H$
$\Delta \overline{V_{Lyman}}=\frac{R_H}{4}$
For Balmer
$\overline{v_{max}}=R_H(\frac{1}{2^2}-\frac{1}{{\infty }^2})=\frac{R_H}{4}$
${\bar{v}}_{min}=R_H(\frac{1}{2^2}-\frac{1}{3^2})=\frac{5}{36}{R}_H$
$\Delta \overline{V_{Balmer}}=\frac{R_H}{4}-\frac{5R_H}{36}=\frac{4R_H}{36}=\frac{RH}{9}$ |
$\frac{\Delta \overline{v_{Lyman}}}{\Delta {\bar{v}}_{Balmer}}=\frac{R_H/4}{R_H/9}=\frac{9}{4}$