Final answer:
The shortest wavelength in the Balmer series for a helium atom is 'x', and the longest wavelength in the Paschen series for a doubly ionized lithium atom is 'x/3', due to the different nuclear charges affecting their spectroscopic series energy levels. The correct option is (c).
Step-by-step explanation:
The student's question pertains to the spectral series of hydrogen-like atoms and involves calculating the wavelengths for different series. To find the shortest wavelength in the Balmer series for a helium ion (He atom), we use the Rydberg formula:
Rydberg formula: 1/λ = RZ2(1/n12 - 1/n22)
For the Balmer series, n1 (the lower energy level) is 2, and the highest n2 (the upper energy level) is ∞, since we're looking for the shortest wavelength. Since He has a nuclear charge (Z) of 2, this would typically result in a spectroscopic series that is four times more energetic than that of hydrogen (where Z=1).
For the Paschen series of the lithium ion (Li2+), n1 is 3 (since Paschen series starts from the third energy level), and the longest wavelength corresponds to the transition where n2 = 4 (the smallest possible energy difference for this series).
The nuclear charge (Z) for the Li2+ ion is 3. When we compare the transitions of the helium and lithium ions accounting for their different nuclear charges, we find that the longest wavelength of the Paschen series for Li2+ is 3 times the shortest wavelength of the Balmer series for He. This is because the energy levels depend on Z2, and when comparing helium and lithium ions, the lithium wavelengths will be (3/2)2 or 9/4 that of helium's, and given that the shortest Balmer wavelength of He is 'x', the longest Paschen wavelength for Li2+ would be 9/4 times 'x' divided by 3 (since we are stepping up from level 3 to 4 in the Paschen series).
Hence, the correct option is C. x/3.