Question

For Balmer series, wavelength of first line is ${\lambda }_1$ and for Brackett series, wavelength of first line is ${\lambda }_2$, then $\frac{{\lambda }_1}{{\lambda }_2}$ is

• A. 0.081
• B. 0.162
• C. 0.198
• D. 0.238

Answer 1

Answer: (B)

0.162

Wavelength for the transition from $n_2$ to $n_1$ is given by $\frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})Z^2$

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

$\therefore$ $\frac{1}{{\lambda }_1}=R(\frac{1}{2^2}-\frac{1}{3^2})Z^2$ $⟹{\lambda }_1=\frac{7.2}{R{Z}^2}$

For first line of Brackett series , $n_1=4$ and $n_2=5$

$\therefore$ $\frac{1}{{\lambda }_2}=R(\frac{1}{4^2}-\frac{1}{5^2})Z^2$ $⟹{\lambda }_2=\frac{44.44}{R{Z}^2}$

$\therefore$ $\frac{{\lambda }_1}{{\lambda }_2}=\frac{7.2}{R{Z}^2}\times \frac{R{Z}^2}{44.44}=0.162$