The ionization energy of hydrogen in excited state is `+0.85 eV`. What will be the energy of the photon emitted when it returns to the ground state?

Explanation

To solve the question, we need to determine the energy of the photon emitted when an electron in a hydrogen atom transitions from an excited state back to the ground state.

### Step-by-Step Solution:

1. **Understand the Ionization Energy**:
The ionization energy of hydrogen in the excited state is given as +0.85 eV. This means that if we provide 0.85 eV of energy to the electron in the excited state, it will be able to escape from the atom.

2. **Know the Ionization Energy of Hydrogen**:
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