Final answer:
The principal quantum number n for a hydrogen atom requiring 0.850 eV to ionize is 4. This is derived from the formula for the energy levels of a hydrogen atom, indicating that the electron is initially in the fourth energy level.The correct answer is d) (4),
Step-by-step explanation:
To determine the principal quantum number (n) for a hydrogen atom that requires 0.850 eV to ionize, we can use the energy formula for the hydrogen atom's electronic states:
En = -13.6 eV/n2
Where En is the energy of the electron at a principal quantum number n, and -13.6 eV is the energy of an electron in the ground state (n=1).
The ionization energy is the energy required to remove the electron from its current state to an infinitely distant point where the energy is 0. We know that the ionization energy is given to be 0.850 eV, and we need to find the corresponding state n for which:
-13.6 eV/n2 = -0.850 eV
By rearranging the formula and solving for n, we find that:
n2 = 13.6 eV / 0.850 eV
n2 ≈ 16
n ≈ 4
Thus, the principal quantum number n when the hydrogen atom requires 0.850 eV to ionize is approximately 4.
The correct answer is d) (4), as the electron must be in the fourth energy level in order to have an ionization energy of 0.850 eV.