The electronic energy of H atom is
`E_n = -(1.312 xx 10^(6))/(n^2) J mol^(-1)`
Calculate
(i) First excitation energy of the electron in the hydrogen atom.
(ii) Ionization energy of the hydrogen atom.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the Energy for the Ground State (n=1)
The electronic energy of the hydrogen atom is given by the formula:
\[ E_n = -\frac{1.312 \times 10^6}{n^2} \text{ J mol}^{-1} \]

For the ground state (n=1):
\[ E_1 = -\frac{1.312 \times 10^6}{1^2} = -1.312 \times 10^6 \text{ J mol}^{-1} \]
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