If the minimum wavelength for H-atom in the Lyman series is 'x', then the maximum wavelength of the Balmer series of He\(^+\) ion in terms of 'x' will be:
The minimum wavelength for the H-atom in the Lyman series (\( \lambda_{\text{min}} \)) is related to the energy transitions in the Lyman series. For the He\(^+\) ion, the energy levels are similar, but the effective nuclear charge is different.
For the H atom (Z = 1), the wavelength in the Lyman series is given by the Rydberg formula:
\[
\frac{1}{\lambda} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right)
\]
where \( R_H \) is the Rydberg constant for hydrogen. Similarly, for the He\(^+\) ion (Z = 2), the Balmer series formula becomes:
\[
\frac{1}{\lambda_{\text{He}^+}} = R_H \cdot Z^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right)
\]
Thus, for He\(^+\), the maximum wavelength of the Balmer series is reduced by a factor of \( \frac{3}{4} \) compared to the wavelength of the H-atom in the Lyman series.
Hence, the correct answer is \( \frac{3}{4} x \).
