A hydrogen atom is in the 4p state.

Determine its energy.

To determine the energy of a hydrogen atom in the 4p state, we can use the energy formula for hydrogen atoms, which is given by:

E = -13.6 eV / n^2

where E is the energy of the atom, -13.6 eV is the ionization energy of hydrogen, and n is the principal quantum number.

In the 4p state, the value of n is 4. Therefore, we can substitute this value into the formula:

E = -13.6 eV / (4^2)
= -13.6 eV / 16
= -0.85 eV

Therefore, the energy of the hydrogen atom in the 4p state is -0.85 eV.

To determine the energy of a hydrogen atom in the 4p state, we can use the Rydberg formula:

E = -13.6 eV / n^2

where E is the energy of the atom, n is the principal quantum number, and -13.6 eV is the ionization energy of hydrogen.

In this case, the principal quantum number is n = 4 because the atom is in the 4p state.

Plugging the values into the formula, we get:

E = -13.6 eV / 4^2
E = -13.6 eV / 16
E = -0.85 eV

So, the energy of the hydrogen atom in the 4p state is approximately -0.85 electron volts (eV).