C

Step 1: Determine the energy levels for He⁺ and Li²⁺.
The energy levels for hydrogen-like ions can be calculated using the formula:
$$ E_n = -\frac{Z^2 \cdot 13.6 \text{ eV}}{n^2} $$
where Z is the atomic number and n is the principal quantum number.

• For He⁺ (Z = 2):
  - E₁ = -54.4 eV
  - E₂ = -13.6 eV
  - E₃ = -6.02 eV
  
• For Li²⁺ (Z = 3):
  - E₁ = -122.4 eV
  - E₂ = -30.6 eV
  - E₃ = -13.6 eV
  

Step 2: Analyze the transitions that can occur.
For He⁺:
The potential excited states are n = 2, 3 (assuming we start from ground state n = 1). The possible transitions are:
- n = 2 to n = 1
- n = 3 to n = 1
- n = 3 to n = 2
This gives us a total of 3 transitions (lines).
For Li²⁺:
The potential excited states are also n = 2, 3. The possible transitions are:
- n = 2 to n = 1
- n = 3 to n = 1
- n = 3 to n = 2
This also provides 3 transitions (lines).
Step 3: Calculate the smallest and largest wavelengths.
Using the formula for the wavelength from the energy gap:
$$ \lambda = \frac{hc}{E} $$
where h is Planck's constant (4.135667696e-15 eV·s) and c is the speed of light (3.00e8 m/s).
For the smallest wavelength, we take the largest energy difference (highest transition):
For He⁺: n=3 to n=1
$$ E = E_1 - E_3 = -54.4 - (-6.02) = 48.38 ext{ eV} $$
$$ \lambda_{smallest} = \frac{(4.135667696 \times 10^{-15} \text{ eV·s})(3.00 \times 10^8 \text{ m/s})}{48.38} \approx 2.57 \times 10^{-12} \text{ m} $$
For the largest wavelength, take the smallest energy transition (lowest):
For He⁺: n=3 to n=2
$$ E = -13.6 - (-6.02) = -7.58 ext{ eV} $$
$$ \lambda_{largest} = \frac{(4.135667696 \times 10^{-15} \text{ eV·s})(3.00 \times 10^8 \text{ m/s})}{7.58} \approx 4.92 \times 10^{-12} \text{ m} $$
The same steps yield similar results for Li²⁺. Thus the answers are:
He⁺: 3 lines, λ = 2.57e-12 m to 4.92e-12 m
Li²⁺: 3 lines, wavelength in the similar range.
Therefore, the answers are (i) 3 lines for both He⁺ and Li²⁺, and for (ii), wavelengths range from smallest to largest: 2.57e-12 m to 4.92e-12 m.