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 $(Z=1)$,we have $n_1 = 1$ and $n_2 = \infty$:$\frac{1}{\lambda_L} = R(1)^2 \left[ \frac{1}{1^2} - \frac{1}{\infty} \right] = R$.Thus,$\lambda_L = \frac{1}{R}$.For the series limit of the Balmer series of a hydrogen-like atom (atomic number $Z$),we have $n_1 = 2$ and $n_2 = \infty$:$\frac{1}{\lambda_B} = R Z^2 \left[ \frac{1}{2^2} - \frac{1}{\infty} \right] = \frac{R Z^2}{4}$.Thus,$\lambda_B = \frac{4}{R Z^2}$.Given that the series limits are equal,$\lambda_L = \lambda_B$:$\frac{1}{R} = \frac{4}{R Z^2} \Rightarrow Z^2 = 4 \Rightarrow Z = 2$.

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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