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What hydrogen like ion has the wavelength difference between the first lines of Balmer and Lyman series equal to 59.3 nm? $R_H = 109678 cm^{-1}$


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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
$\large\frac{1}{\lambda_2} = R_H.Z^2[\large\frac{1}{1^2}-\large\frac{1}{2^2}$]
=$\large\frac{3}{4}\times R_1\times Z^2$
$\lambda_2 = \large\frac{4}{3R_H.Z^2}$
$\lambda_B-\lambda = 59.3\times10^{-7} $cm
$\large\frac{36}{5R_HZ^2} - \large\frac{4}{3R_H.Z^2} = 59.3\times10^{-7}$
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)
answered Jan 31, 2014 by sharmaaparna1

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