Energy of electrons in the first excited state of hydrogen:
The energy of an electron in the first excited state of hydrogen is given as -3.4 electron volt (eV). This value represents the total energy of the electron in this particular energy level.

Calculating the kinetic energy (KE) of the electron in the ground state:
The ground state of hydrogen is the lowest energy level that an electron can occupy. In this state, the electron is closest to the nucleus and has the lowest energy. To calculate the kinetic energy (KE) of the electron in the ground state, we need to use the formula:

KE = Total Energy - Potential Energy

Since the electron in the first excited state has a total energy of -3.4 eV, we can substitute this value into the equation:

KE = -3.4 eV - PE

Calculating the potential energy (PE) of the electron in the ground state:
The potential energy (PE) of the electron in the ground state can be calculated by rearranging the above equation:

PE = -3.4 eV - KE

Substituting values to find KE and PE:
Since the electron in the ground state has the lowest energy, its kinetic energy (KE) will be the lowest possible value. We can assume that the kinetic energy is zero, as the electron is not moving in the ground state. Let's substitute this value into the equation:

KE = -3.4 eV - 0 eV
KE = -3.4 eV

Therefore, the kinetic energy of the electron in the ground state is -3.4 eV.

To find the potential energy (PE) of the electron in the ground state, we can now substitute the calculated value of KE into the equation:

PE = -3.4 eV - (-3.4 eV)
PE = -3.4 eV + 3.4 eV
PE = 0 eV

Therefore, the potential energy of the electron in the ground state is 0 eV.

Conclusion:
In summary, the kinetic energy (KE) of the electron in the ground state of hydrogen is -3.4 eV, while the potential energy (PE) is 0 eV. This means that the electron in the ground state has no movement (zero kinetic energy) but still has a potential energy due to its position relative to the nucleus.