Question

A hydrogen atom in the ground state absorbs a photon of wavelength 93.8 nm. what energy level does the electron reach? this excited atom then emits a photon of wavelength 1094.0 nm. what energy level does the electron fall to?

Answer 1

When a hydrogen atom absorbs a photon, its electron is excited to a higher energy level. The electron in the hydrogen atom reaches the energy level n = 7 when it absorbs a photon with a wavelength of 93.8 nm. The electron then falls to the energy level n = 16 when the atom emits a photon with a wavelength of 1094.0 nm.

Step-by-step explanation:

When a hydrogen atom absorbs a photon, the electron in the atom is excited to a higher energy level. In this case, the hydrogen atom in the ground state absorbs a photon with a wavelength of 93.8 nm. The energy of the photon can be calculated using the equation E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength.

Substituting the values, we get E = (6.626 × 10^-34 J.s × 3 × 10^8 m/s) / (93.8 × 10^-9 m) = 2.107 × 10^-18 J. To determine the energy level the electron reaches, we can use the equation for the energy of an electron in the hydrogen atom: E = -13.6 eV / n^2, where n is the principal quantum number. Rearranging the equation, we can solve for n: n^2 = -13.6 eV / E and n = √(-13.6 eV / 2.107 × 10^-18 J) = 7.01. Therefore, the electron reaches the energy level n = 7.

Afterwards, the excited hydrogen atom emits a photon with a wavelength of 1094.0 nm. The energy of this photon can also be calculated using the same equation E = hc/λ, giving us E = (6.626 × 10^-34 J.s × 3 × 10^8 m/s) / (1094.0 × 10^-9 m) = 1.811 × 10^-19 J. We can follow the same process as before to determine the energy level the electron falls to. We solve the equation n^2 = -13.6 eV / E for n: n = √(-13.6 eV / 1.811 × 10^-19 J) = 15.9. Therefore, the electron falls to the energy level n = 16.

Answer 2

Answer: So you can use the Bohr formula En = -13.6*(1/n2^2 - 1/n1^2) eV where En is in electron volts. Energy of a photon is Ep = 1.24/wl eV where wl is wavelength in microns So 93.8 nm = 0.0938 um so Ep = 13.22 ev. Now atom radiates a photon wl' = 7.459 um loosing an energy E' of E' = 1.24/7.459 = 0.166 eV SO the atom has a net gain in energy of Ep - E' =13.053 eV TO find th eenergy level, set Ep - E' = En 13.053 = -13.6 (1/n2^2 - 1/n1^2) and set n1 = 1 = ground state. then solve for n2 -13.055/13.6 = 1/n2^2 - 1 --> 1 - 13.055/13.6 = 1/n2^2 0.040193 = 1/n2^2 ---> n2 = sqrt(1/0.040193) = 4.98 ~ 5 So n final is 5