A hydrogen atom is in its n = 2 excited state when its electron absorbs a photon of energy 8.5 eV. What is the energy of the resulting free electron? The lowest level energy state of hydrogen is -13.6 eV. (h = 6.626 × 10^-34 J ∙ s, 1 eV = 1.60 × 10^-19 J) A) 5.1 eV B) 6.6 eV C) 6.9 eV D) 7.6 eV

Question

Quizplus · Accepted Answer

First, we need to find the final energy state of the hydrogen atom after the electron absorbs the photon. We can use the formula:

ΔE = hf = (E2 - E1)

where ΔE is the change in energy, h is Planck's constant, f is the frequency of the photon, and E2 and E1 are the energy levels before and after the absorption, respectively.

We know that the atom is in its n=2 excited state before the absorption, which corresponds to an energy level of:

E2 = -13.6 eV / 2² = -3.4 eV

We also know the energy of the absorbed photon, which is given as 8.5 eV. Therefore:

ΔE = hf = 8.5 eV

So, we can solve for the final energy level of the electron:

E1 = E2 + ΔE = -3.4 eV + 8.5 eV = 5.1 eV

Finally, we need to find the energy of the resulting free electron. Since the electron is now free (i.e. not bound to the hydrogen nucleus), its energy is simply the same as the final energy level we just calculated:

5.1 eV

Therefore, the answer is A) 5.1 eV.

A Hydrogen Atom Is in Its N = 2 Excited