$(a)\;Mg^{2+}\qquad(b)\;Na^{1+}\qquad(c)\;Li^{+2}\qquad(d)\;K^{2+}$

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We have $\large\frac{1}{\lambda} = R_HZ^2[\large\frac{1}{n_1^2}-\large\frac{1}{n_2^2}$]

Balmer series

$\large\frac{1}{\lambda_B} = R_H Z^2[\large\frac{1}{2^2}-\large\frac{1}{3^2}]$

$=\large\frac{5}{36}\times R_H\times Z^2$

$\lambda_B = \large\frac{36}{SR_HZ^2}$ ----(1)

Lyman Series

$\large\frac{1}{\lambda_2} = R_H.Z^2[\large\frac{1}{1^2}-\large\frac{1}{2^2}$]

=$\large\frac{3}{4}\times R_1\times Z^2$

$\lambda_2 = \large\frac{4}{3R_H.Z^2}$

Given

$\lambda_B-\lambda = 59.3\times10^{-7} $cm

$\large\frac{36}{5R_HZ^2} - \large\frac{4}{3R_H.Z^2} = 59.3\times10^{-7}$

$\large\frac{1}{R_H.Z^2}[7.2-1.333]=59.3\times10^{-7}$

or $Z^2 = \large\frac{5.867}{R_H\times 59.3\times10^{-7}}$

=$\large\frac{5.867}{109678\times 59.3\times10^{-7}}$

$\therefore$ Z = 3

$\therefore$ H like atom is $Li^{2+}$

Hence answer is (c)

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