Statement I : The energy of a `He^(+)` ion for a given `n` is almost exactly four times that of H atom for the same `n`.
Statement II : Photon emitted during transition between corresponds pair of levels in `He^(+)` and H have the same energy `E` and the same wavelength `lambda = hc//E`.

Options

A. Statement I is True , Statement II is True , Statement II is a correct explanation for Statement I.
B. Statement I is True , Statement II is True , Statement II is NOT a correct explanation for Statement I.
C. Statement I is True , Statement II is False.
D. Statement I is False , Statement II is True.

Explanation

To solve the question, we need to analyze both statements regarding the energy levels of the `He^(+)` ion and the hydrogen atom.

**Step 1: Analyze Statement I**
- The energy of an electron in a hydrogen atom is given by the formula:
\[ E = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.
- For a helium ion `He^(+)`, which has a nuclear charge \( Z = 2 \), the energy is given by:
\[ E_{He^+} = -\frac{13.6 \times Z^2 \, \text{eV}}{n^2} = -\frac{13.6 \times 2^2 \, \text{eV}}{n^2} = -\frac{54.4 \, \text{eV}}{n^2} \]
- Thus, the energy of the `He^(+)` ion is:
\[ E_{He^+} = 4 \times \left(-\frac{13.6 \, \text{eV}}{n^2}\right) = 4 \times E_{H} \]
- Therefore, Statement I is **true**.

**Step 2: Analyze Statement II**
- The energy of the photon emitted during a transition between two energy levels is given by the difference in energy levels:
\[ E = E_{initial} - E_{final} \]
- For both hydrogen and `He^(+)`, if we consider transitions between corresponding energy levels (with the same \( n \)), the energy emitted will be proportional to \( Z^2 \):
- For hydrogen \( (Z=1) \):
\[ E_H = -\frac{13.6 \, \text{eV}}{n^2} \]
- For `He^(+)` \( (Z=2) \):
\[ E_{He^+} = -\frac{54.4 \, \text{eV}}{n^2} \]
- The energy of the photon emitted from `He^(+)` is four times that from hydrogen, hence:
\[ E_{He^+} = 4 \times E_H \]
- The wavelength \( \lambda \) of the emitted photon is given by:
\[ \lambda = \frac{hc}{E} \]
- Since the energy of the photon from `He^(+)` is four times that of hydrogen, the wavelength will be inversely proportional to the energy:
\[ \lambda_{He^+} = \frac{hc}{4E_H} = \frac{1}{4} \lambda_H \]
- Therefore, the wavelengths are not the same, making Statement II **false**.

**Final Conclusion:**
- Statement I is true, and Statement II is false.