Question

The ratio of the frequency corresponding to the third line in Lyman series of hydrogen atomic spectrum to that of the first line in Balmer series of ${Li}^{2+}$ spectrum is:

• A. $\frac{4}{5}$
• B. $\frac{5}{4}$
• C. $\frac{4}{3}$
• D. $\frac{3}{4}$
• E. $\frac{3}{8}$

Answer 1

The correct option is D $\frac{3}{4}$

For Lyman series, $n_1=1$
and for the third line, $n_2=4$

${\overline{v}}_1=R_H{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

$=R_H{\left(1\right)}^2(\frac{1}{1}-\frac{1}{16})=\frac{15}{16}{R}_H$ ($\because$ For $H$ atom $Z=1$)

For Balmer series, $n_1=2$ and for first line, $n_2=3$

${\overline{v}}_2=R_H{Z}^2(\frac{1}{{\left(2\right)}^2}-\frac{1}{{\left(3\right)}^2})$ $[For{Li}^{2+},Z=3]$

$=R_H{\left(3\right)}^2(\frac{1}{4}-\frac{1}{9})=9R_H\left(\frac{5}{36}\right)$

$\frac{v_1}{v_2}=\frac{15R_H\times 36}{16\times 9\times 5R_H}=\frac{3}{4}$