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 Li²⁺ spectrum is

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

Answer 1

Answer: (D)

For Lyman series, $n_{1}=1$

and for third line, $n_{2}=4$

$\bar{v}_{1} =R_{ H } Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$

$=R_{ H }(1)^{2}\left(\frac{1}{1}-\frac{1}{16}\right)=\frac{15}{16} R_{ H }$

$(\because$ For $H$ atom $Z=1)$

For Balmer series, $n_{1}=2$ and for first line, $n_{2}=3$

$\bar{v}_{2} =R_{ H } Z^{2}\left(\frac{1}{(2)^{2}}-\frac{1}{(3)^{2}}\right) \left[\text { For } Li ^{2+}, Z=3\right] $

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

$\frac{v_{1}}{v_{2}} =\frac{15 R_{ H } \times 36}{16 \times 9 \times 5 R_{ H }}=\frac{3}{4}$