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

Answer: 3

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 ${\lambda }_B=\frac{36^2}{5R_H{Z}^2}$

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 ${\lambda }_B-{\lambda }_L=59.3\times 10^{-7}=\frac{36}{5R_H{Z}^2}-\frac{4}{3R_H{Z}^2}$

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

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

or $Z=3$

Hydrogen-like species is $L{i}^{2+}$