Hydrogen spectrum

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If the series limit of Lymen series for Hydrogen atom is equal to the series limit of Balmer series for a hydrogen like atom, then what is the atomic number of this hydrogen like atom?

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Solution

$By using \frac{1}{λ} = R Z^{2} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$
$For hydrogen atom \frac{1}{{(λ_{min})}_{H}} = R [\frac{1}{1^{2}} - \frac{1}{\infty}] = R$
$\Rightarrow {(λ_{min})}_{H} = \frac{1}{R}$
$For hydrogen like atom {(\frac{1}{λ_{min}})}_{atom} = R Z^{2} (\frac{1}{2^{2}} - \frac{1}{\infty})$
$\Rightarrow {(λ_{min})}_{atom} = \frac{4}{R Z^{2}}$
$From Eqs. (i) and (ii) \frac{1}{R} = \frac{4}{R Z^{2}} \Rightarrow Z = 2 .$

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Hydrogen spectrum

Remember concepts with our Masterclasses.

If the series limit of Lymen series for Hydrogen atom is equal to the series limit of Balmer series for a hydrogen like atom, then what is the atomic number of this hydrogen like atom?

By using 1λ=RZ21n12−1n22 For hydrogen atom 1λminH=R112−1∞=R⇒ λminH=1R For hydrogen like atom 1λminatom =RZ2122−1∞⇒ λminatom=4RZ2 From Eqs. (i) and (ii) 1R=4RZ2⇒Z=2 .

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