A hydrogen atom is in a state of ionization energy 0.85 eV. If it makes a transition to the ground state, what is the energy of the emitted photon.

Explanation

To find the energy of the emitted photon when a hydrogen atom transitions from an excited state to the ground state, we can follow these steps:

### Step 1: Understand the Energy States
The hydrogen atom is in an excited state with an ionization energy of 0.85 eV. This means that the total energy of the atom in this state is -0.85 eV (since ionization energy is defined as the energy required to remove an electron from the atom, and thus is positive).

### Step 2: Determine the Ground State Energy
The energy of the hydrogen atom in the ground state (n=1) can be calculated using the formula:
\[
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