Question

The ratio of difference in wavelengths of 1st and 2nd lines of Lyman series in H-like atom to difference in wavelength for 2nd and 3rd lines of same series is:

• A. 2.5:1
• B. 3.5:1
• C. 4.5:1
• D. 5.5:1

Answer 1

Answer: (B)

3.5:1

The formula for wavelength of Lyman series is written as

$\frac{1}{\lambda }=R[1-\frac{1}{n^2}]$

For first, second and third lines of Lyman series $n=2,3,4$ respectively.

For first-line of Lyman series

$\frac{1}{{\lambda }_1}=R[1-\frac{1}{2^2}]=\frac{3R}{4}$

$\Rightarrow {\lambda }_1=\frac{4}{3R}$

For second-line of Lyman series

$\frac{1}{{\lambda }_2}=R[1-\frac{1}{3^2}]=\frac{8R}{9}$

$\Rightarrow {\lambda }_2=\frac{9}{8R}$

For the third line of Lyman series

$\frac{1}{{\lambda }_3}=R[1-\frac{1}{4^2}]=\frac{15R}{16}$

$\Rightarrow {\lambda }_3=\frac{16}{15R}$

Difference between $1^{st}$ and $2^{nd}$ line wavelength

${\lambda }_1-{\lambda }_2=\frac{4}{3R}-\frac{9}{8R}=\frac{5}{24R}$

Difference between $2^{nd}$ and $3^{rd}$ line wavelength

${\lambda }_2-{\lambda }_3=\frac{9}{8R}-\frac{16}{15R}=\frac{7}{120R}$

Ratio = $\frac{{\lambda }_1-{\lambda }_2}{{\lambda }_2-{\lambda }_3}=\frac{25}{7}=3.5:1$

Hence, the correct option is $B$