If wavelength of the 4th line of balmer series of He+ ion is X, then calculate the wavelength of 3rd line of balmer series of Be3+ ion in terms of X?

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Solution

The Rydberg equation is given as: $v = \frac{1}{λ} = R z^{2} (\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}})$
For the fourth line in the Balmer series, the initial and final states are 6 and 2 respectively.
Hence, the wavelength of the fourth line of the Balmer series of $H e^{+}$ ion is:
$v = \frac{1}{λ} = R z^{2} (\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}})$
$= R 1^{2} (\frac{1}{2^{2}} - \frac{1}{6^{2}})$
$= R (\frac{1}{4} - \frac{1}{36}) = \frac{8}{36} R$
$∴ λ = \frac{36}{8 R} = \frac{4.5}{R} = X$

The wavelength of the third line of the Balmer series of $B e^{3 +}$ ion is:
$v = \frac{1}{λ} = R z^{2} (\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}})$
$= R 3^{2} (\frac{1}{2^{2}} - \frac{1}{5^{2}})$
$= 9 R (\frac{1}{4} - \frac{1}{25}) = \frac{189}{100} R$
$∴ λ = \frac{100}{189 R} = \frac{0.5291}{R}$
Therefore, the wavelength of the third line of the Balmer series of $B e^{3 +}$ ion in terms of X is :
$λ_{H e^{+}} = \frac{4.5}{R} = X$
$λ_{B e^{3 +}} = \frac{0.5291}{R} = y t i m e s o f X$
$∴ \frac{X}{y X} = \frac{4.5}{0.5291} = 8.5$
$∴ y = \frac{1}{8.5} X$

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