In which hydrogen-like ion is the difference | vedclass

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(A) For the first line of the Balmer series $(n_1 = 2, n_2 = 3)$:$\frac{1}{\lambda_B} = Z^2 R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = Z^2 R_H

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$Li^{2+}$

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(A) For the first line of the Balmer series $(n_1 = 2, n_2 = 3)$:$\frac{1}{\lambda_B} = Z^2 R_H \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = Z^2 R_H \left( \frac{5}{36} ight) \implies \lambda_B = \frac{36}{5 Z^2 R_H}$For the first line of the Lyman series $(n_1 = 1, n_2 = 2)$:$\frac{1}{\lambda_L} = Z^2 R_H \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = Z^2 R_H \left( \frac{3}{4} ight) \implies \lambda_L = \frac{4}{3 Z^2 R_H}$Given the difference $\lambda_B - \lambda_L = 59.3 \ nm = 59.3 	imes 10^{-7} \ cm$:$\frac{36}{5 Z^2 R_H} - \frac{4}{3 Z^2 R_H} = 59.3 	imes 10^{-7}$$\frac{1}{Z^2 R_H} \left( \frac{108 - 20}{15} ight) = 59.3 	imes 10^{-7}$$\frac{88}{15 Z^2 (109678)} = 59.3 	imes 10^{-7}$$Z^2 = \frac{88}{15 	imes 109678 	imes 59.3 	imes 10^{-7}} \approx 9$$Z = 3$. Thus,the ion is $Li^{2+}$.

In which hydrogen-like ion is the difference in wavelength between the first line of the Balmer series and the first line of the Lyman series equal to $59.3 \ nm$? $[R_H = 109678 \ cm^{-1}]$

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