If the series limit of the Lyman series for a | vedclass

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(B) The Rydberg formula for the wavelength of emitted radiation 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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$2$

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(B) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.For the series limit of the Lyman series of a Hydrogen atom $(Z=1)$: Here $n_1 = 1$ and $n_2 = \infty$.$\frac{1}{\lambda_{Lyman}} = R(1)^2 \left[ \frac{1}{1^2} - \frac{1}{\infty} \right] = R$.For the series limit of the Balmer series of a hydrogen-like atom with atomic number $Z$: Here $n_1 = 2$ and $n_2 = \infty$.$\frac{1}{\lambda_{Balmer}} = R Z^2 \left[ \frac{1}{2^2} - \frac{1}{\infty} \right] = \frac{R Z^2}{4}$.Given that the series limits are equal,$\lambda_{Lyman} = \lambda_{Balmer}$,which implies $\frac{1}{\lambda_{Lyman}} = \frac{1}{\lambda_{Balmer}}$.$R = \frac{R Z^2}{4}$.$1 = \frac{Z^2}{4} \Rightarrow Z^2 = 4 \Rightarrow Z = 2$.

If the series limit of the Lyman series for a Hydrogen atom is equal to the series limit of the Balmer series for a hydrogen-like atom,then the atomic number of this hydrogen-like atom will be:

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