If the shortest wavelength of hydrogen atom in the Lyman series is \( x \), then the longest wavelength in the Balmer series of He\(^+\) is:
Detailed Step-by-Step Solution
This problem is solved using the Rydberg Formula:
Step 1: Shortest Wavelength of H in Lyman Series
For the hydrogen atom, \( Z = 1 \).
In the Lyman series, the electron falls to \( n_1 = 1 \). The shortest wavelength corresponds to the maximum energy transition, which occurs when the electron falls from infinity (\( n_2 = \infty \)).
\( \frac{1}{x} = R_H (1 - 0) \)
\( R_H = \frac{1}{x} \quad \text{--- (Equation 1)} \)
Step 2: Longest Wavelength of He\(^+\) in Balmer Series
For the Helium ion (He\(^+\)), the atomic number is \( Z = 2 \).
In the Balmer series, the electron falls to \( n_1 = 2 \). The longest wavelength corresponds to the minimum energy transition, which occurs when the electron falls from the immediately adjacent higher shell (\( n_2 = 3 \)).
Let this wavelength be \( \lambda' \):
\( \frac{1}{\lambda'} = R_H \cdot 4 \left( \frac{1}{4} - \frac{1}{9} \right) \)
Solve the fraction inside the brackets:
\( \frac{1}{\lambda'} = \frac{5 R_H}{9} \)
Step 3: Substitute and Find \(\lambda'\)
From Equation 1, we know that \( R_H = \frac{1}{x} \). Substitute this into our result from Step 2:
\( \frac{1}{\lambda'} = \frac{5}{9x} \)
Now, invert the equation to find the wavelength \( \lambda' \):
Conclusion: The longest wavelength in the Balmer series of He\(^+\) is exactly \( \frac{9x}{5} \). Therefore, the correct option is (A).
Cracking Atomic Spectrum Numericals
This question tests two highly critical concepts simultaneously: the effect of atomic number (\(Z^2\)) on hydrogen-like species, and the difference between "longest" (minimum energy gap, e.g., $3 \to 2$) and "shortest" (maximum energy gap, e.g., $\infty \to 1$) wavelength transitions.
To solidify your understanding of emission spectra, Rydberg equations, and the Bohr Model, we highly recommend reading our detailed chapter notes on the Structure of Atom Class 11 Chemistry.
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