An excited `He^(+)` ion emits two photons in succession, with wavelength 108.5 nm and 30.4 nm, in making a transition to ground state. The quantum number n, corresponding to its initial excited state is (for photon of wavelength `lamda`, energy `E=(1240eV)/(lamda("in nm"))`

Explanation

To solve the problem, we need to find the initial quantum number \( n \) of the excited helium ion \( \text{He}^+ \) that emits two photons with given wavelengths. The steps are as follows:

### Step 1: Calculate the energy of each photon
The energy \( E \) of a photon can be calculated using the formula:
\[ E = \frac{1240 \, \text{eV}}{\lambda \, (\text{in nm})} \]
We will calculate the energy for both wavelengths.
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