Chemistry - Atomic Structure Question with Solution | TestHub

Wavelength of 1st line in Balmer series,
$\frac{\text{1}}{{\text{λ}}_{\text{B}}}\text{=}{\text{Z}}^{\text{2}}{\text{R}}_{\text{H}}\left[\frac{\text{1}}{{\text{2}}^{\text{2}}}-\frac{\text{1}}{{\text{3}}^{\text{2}}}\right]\text{=}\frac{\text{5}}{\text{36}}{\text{R}}_{\text{H}}{\text{Z}}^{\text{2}}$
or${\text{λ}}_{\text{B}}=\frac{36}{5{\text{R}}_{\text{H}}{\text{Z}}^{\text{2}}}$
Wavelength of 1st line in Lyman series is,$\frac{1}{{\text{λ}}_{\text{L}}}={\text{Z}}^{\text{2}}{\text{R}}_{\text{H}}\left[\frac{1}{1^2}-\frac{1}{2^2}\right]$
or${\text{λ}}_{\text{L}}=\frac{4}{3\times {\text{R}}_{\text{H}}{\text{Z}}^2}$
Difference${\text{λ}}_{\text{B}}-{\text{λ}}_{\text{L}}=59.3\times {10}^{-7}=\frac{36}{5{\text{R}}_{\text{H}}{\text{Z}}^{\text{2}}}-\frac{4}{3{\text{R}}_{\text{H}}{\text{Z}}^{\text{2}}}$
$=\frac{1}{{\text{R}}_{\text{H}}{\text{Z}}^{\text{2}}}\left[\frac{36}{5}-\frac{4}{3}\right]$
${\text{Z}}^{\text{2}}=\frac{88}{59.3\times {10}^{-7}\times 109678\times 15}=9.0$
or$\text{Z}=3$
Hydrogen-like species is${\text{Li}}^{\text{2+}}$