The electron in a hydrogen atom, originally in level n = 8, undergoes a transition to a lower level by emitting a photon of wavelength 3745 nm. What is the final level of the electron? a) 5, b)8, c)9, d)6
To determine the final level of the electron in a hydrogen atom, we can use the equation for the energy of a photon (E) emitted during a transition:
E = hc/λ,
where h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (2.998 x 10^8 m/s), and λ is the wavelength of the photon.
First, let's convert the wavelength of the photon from nanometers (nm) to meters (m):
3745 nm = 3745 x 10^-9 m.
Now, we can substitute the values into the equation to find the energy of the photon:
E = (6.626 x 10^-34 J·s) * (2.998 x 10^8 m/s) / (3745 x 10^-9 m).
Calculating this expression will give us the energy of the photon.
The energy difference between two energy levels in the hydrogen atom is given by the equation:
ΔE = -13.6 eV * (1/n_final^2 - 1/n_initial^2),
where n_final is the final energy level, n_initial is the initial energy level, and -13.6 eV is the ionization energy of hydrogen.
To find the final level, we can rearrange the equation to solve for n_final and substitute the values:
ΔE = -13.6 eV * (1/n_final^2 - 1/8^2) = E.
Solving this equation will give us the value of n_final.
Once we have calculated n_final, we can compare it to the provided answer choices (a, b, c, d) to determine the correct answer.