Let the potential energy of hydrogen atom in the ground state be zero. Then its total energy in the first excited state will be

To determine the energy of a hydrogen atom in the first excited state, we will consider the potential energy of hydrogen atom is zero in the ground state, as stated in the problem. Let's break this down step by step:

1. Understand the Ground State Energy:

The energy of a hydrogen atom in the ground state (n = 1) is given by the formula:

En = -13.6 eV / n2

For the ground state (n = 1), the energy is:

E1 = -13.6 eV / (1)2 = -13.6 eV

2. Set the Potential Energy to Zero:

The problem specifies that the potential energy of hydrogen atom is zero in the ground state. This means we need to adjust the total energy to account for this condition. The total energy is a sum of kinetic energy and potential energy:

Total Energy = Kinetic Energy + Potential Energy

If the potential energy of hydrogen atom is zero, the total energy becomes:

Total Energy = 0 + (-13.6 eV) = 13.6 eV

However, we know the actual total energy in the ground state is:

Etotal = -13.6 eV (Total Energy in Ground State)

3. Calculate the Energy in the First Excited State:

Next, we calculate the energy of the hydrogen atom in the first excited state (n = 2). The formula for the energy in any nth state is:

En = -13.6 eV / n2

For the first excited state (n = 2), the energy is:

E2 = -13.6 eV / (2)2 = -13.6 eV / 4 = -3.4 eV

4. Adjust for the New Reference Point:

Since we have set the potential energy of hydrogen atom is zero in the ground state, we need to adjust the energy in the excited state as well. This can be done by adding the energy difference between the ground state and the first excited state to the reference total energy, which is 27.2 eV (calculated earlier). Thus:

E2 (new) = 27.2 eV - 3.4 eV = 23.8 eV

5. Final Answer:

Therefore, the energy of the hydrogen atom in the first excited state, when the potential energy of hydrogen atom is set to zero in the ground state, is 23.8 eV.