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(B) The Rydberg formula for the wavelength of a spectral line is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

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$Be^{3+}$

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(B) The Rydberg formula for the wavelength of a spectral line is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For the limiting line of the Lyman series,$n_1 = 1$ and $n_2 = \infty$,so $\frac{1}{\lambda} = R Z^2$.Let $\lambda_H$ be the wavelength for the hydrogen atom $(Z_H = 1)$ and $\lambda_X$ be the wavelength for the ion $X$ $(Z_X = Z)$.Given $\lambda_H = 16 \lambda_X$,we have $\frac{1}{\lambda_H} = \frac{1}{16} \frac{1}{\lambda_X}$.Substituting the Rydberg formula: $R(1)^2 = \frac{1}{16} R(Z)^2$.$1 = \frac{Z^2}{16} \implies Z^2 = 16 \implies Z = 4$.The ion with atomic number $Z = 4$ is $Be^{3+}$.

If the wavelength of the limiting line of the Lyman series of the hydrogen atom is $16$ times the wavelength of the limiting line of the same series for an ion $X$,then what is $X$?

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