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(A) The wavelength $\lambda$ for a hydrogen-like atom is given by the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_

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$2$

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(A) The wavelength $\lambda$ for a hydrogen-like atom is given by the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For the shortest wavelength of the Lyman series of hydrogen $(Z=1)$,$n_1 = 1$ and $n_2 = \infty$. Thus,$\frac{1}{\lambda_L} = R (1)^2 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R$.For the shortest wavelength of the Balmer series of a hydrogen-like atom $(Z)$,$n_1 = 2$ and $n_2 = \infty$. Thus,$\frac{1}{\lambda_B} = R Z^2 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R \frac{Z^2}{4}$.Given that $\lambda_L = \lambda_B$,we have $\frac{1}{\lambda_L} = \frac{1}{\lambda_B}$.Therefore,$R = R \frac{Z^2}{4}$,which implies $Z^2 = 4$.Solving for $Z$,we get $Z = 2$.

The shortest wavelength of the Lyman series of a hydrogen atom is equal to the shortest wavelength of the Balmer series of a hydrogen-like atom of atomic number $Z$. The value of $Z$ is equal to:

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