To determine the energy of the radiation emitted when an electron in hydrogen transitions from the
3‌rd ‌ excited state to the ground state, first, we need to understand what is meant by "ground state" and "excited states" in the context of a hydrogen atom.
The energy levels of an electron in a hydrogen atom are given by the formula:
En=−‌where
En is the energy of the n-th energy level (with the ground state being
n=1 ) and
13.6‌eV is the energy of the electron in the ground state. The ground state
(n=1) has an energy of
−13.6‌eV. The
3‌rd ‌ excited state would correspond to
n=4, since the excited states start counting after the ground state
n=1 (first excited state is
n=2, second is
n=3, and so forth).
Therefore, the energy of the
3‌rd ‌ excited state
(n=4) is:
To find the energy of the radiation emitted when the electron transitions from the
n=4 level to the ground state (
n=1 ), we calculate the difference in energy between these two levels:
∆E=E‌final ‌−E‌initial ‌=E1−E4=(−13.6‌eV)−(−0.85‌eV)=−13.6‌eV+0.85‌eV=−12.75‌eV
The negative sign indicates that energy is released, so the magnitude of the energy is
12.75‌eV. Thus, the energy of the radiation emitted is
12.75‌eV.
This corresponds to Option B.