Question

Last line of Lyman series for H-atom has wavelength ${\lambda }_1\mathring{A}$ The $2^{\text{nd }}$ line of Balmer series has wavelength ${\lambda }_2\mathring{A},$ then:

• A. $\frac{16}{{\lambda }_1}=\frac{9}{{\lambda }_2}$
• B. $\frac{16}{{\lambda }_2}=\frac{3}{{\lambda }_1}$
• C. $\frac{4}{{\lambda }_1}=\frac{1}{{\lambda }_2}$
• D. $\frac{16}{{\lambda }_1}=\frac{3}{{\lambda }_2}$

Answer 1

Answer: (B)

Last line of lyman series refers to transition $\infty \to 1$

$\frac{1}{{\lambda }_1}=R\left(\frac{1}{1^2}-\frac{1}{{\infty }^2}\right)$

$\frac{1}{{\lambda }_1}=R...(1)$

$2^{nd}$ line of Balmer series refers to transition $4\to 2$

$\frac{1}{{\lambda }_2}=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right)$

$\frac{1}{{\lambda }_2}=R\left(\frac{1}{4}-\frac{1}{16}\right)$

$\frac{1}{{\lambda }_2}=R\times \frac{3}{16}...(2)$

$\frac{\frac{1}{{\lambda }_1}}{\frac{1}{{\lambda }_2}}$

$=\frac{R}{\frac{3R}{16}}=\frac{R}{1}\times \frac{16}{3R}$

$=\frac{{\lambda }_2}{{\lambda }_1}=\frac{16}{3}$