Question

If wavelength corresponding to $2^{nd}$ line of Lyman series is $\lambda$, then the wavelength corresponding to last line of Balmer series will be:

• A. $\frac{9 \lambda}{8}$
• B. $\frac{32 \lambda}{9}$
• C. $\frac{27 \lambda}{16}$
• D. $\frac{24 \lambda}{7}$

Answer 1

Answer: (B)

$\frac{1}{\lambda}=R Z^{2}\left(\frac{1}{1^{2}}-\frac{1}{3^{2}}\right)$
$=\frac{8 R Z^{2}}{9}$
$\frac{1}{\lambda^{\prime}}=R Z^{2}\left(\frac{1}{2^{2}}-\frac{1}{(\infty)^{2}}\right)$
$=\frac{R Z^{2}}{4}$
$\frac{1}{\lambda'}=\frac{9}{32 \lambda}$
$\lambda'=\frac{32 \lambda}{9}$