Question

Atomic number of Hydrogen like species that has a wave length difference of 59.3 nm between first lin $e$ of Balmer and first line of Lyman series is $\left(R_H=109678 \mathrm{~cm}^{-1}\right)$

Answer 1

Wavelength of 1st line in Balmer series, $\frac{1}{\lambda_{ B }}= Z ^{2} R _{ H }\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right]=\frac{5}{36} R _{ H } Z ^{2}$

or $\lambda_{ B }=\frac{36^{2}}{5 R _{ H } Z ^{2}}$

Wavelength of ist line in Lyman series is, $\frac{1}{\lambda_{ L }}= Z ^{2} R _{ H }\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}\right]$

or $\lambda_{ L }=\frac{4}{3 \times R _{ H } Z ^{2}}$

Difference $\lambda_{ B }-\lambda_{ L }=59.3 \times 10^{-7}=\frac{36}{5 R _{ H } Z ^{2}}-\frac{4}{3 R _{ H } Z ^{2}}$

$=\frac{1}{ R _{ H } Z ^{2}}\left[\frac{36}{5}-\frac{4}{3}\right]$

$Z^{2}=\frac{88}{59.3 \times 10^{-7} \times 109678 \times 15}=9.0$

or $Z=3$

Hydrogen-like species is $Li ^{2+}$