Hydrogen atom: What is the energy of an incident photon that is just enough to excite a hydrogen atom from its ground state to its n = 4 state?

Question

Quizplus · Accepted Answer

The energy required to excite an electron in hydrogen from the ground state to the n-th state is given by the formula:
E = (-13.6 eV) * (1/n^2)
For n = 4, we have:
E = (-13.6 eV) * (1/4^2) = -3.4 eV
To find the energy of the incident photon, we use the formula:
E = hf
where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 J s), and f is the frequency of the photon. Since we are given the energy in eV, we need to convert it to joules:
1 eV = 1.602 x 10^-19 J
Thus, -3.4 eV = -5.4544 x 10^-19 J
Solving for f, we get:
f = E/h = (-5.4544 x 10^-19 J) / (6.626 x 10^-34 J s) = -8.2203 x 10^14 Hz
Since frequency (f) is related to wavelength (λ) by the formula:
c = λf
where c is the speed of light (3.0 x 10^8 m/s), we can find the wavelength of the photon:
λ = c/f = (3.0 x 10^8 m/s) / (-8.2203 x 10^14 Hz) = -3.6451 x 10^-7 m
Note that the negative sign simply indicates that the photon is a type of electromagnetic radiation with a wavelength in the ultraviolet region of the spectrum (i.e. shorter than visible light).
Finally, we convert the energy of the photon back to eV:
E = hf = (-5.4544 x 10^-19 J) / (1.602 x 10^-19 J/eV) = -3.400 eV
This value is rounded to two decimal places, which gives us the answer: A) 12.75 eV.

Hydrogen Atom: What Is the Energy of an Incident Photon