Question

An element is recently discovered with ionization energy of 20eV. This element has excited states that have energies of 13eV, 15eV, 17.5eV and 19.2eV above ground state energy.

A) draw an energy level diagram, include ground state and all of exited states.

B) If 17.5eV photon is absorbed by an atom of this substance, in which states was the atom before it absorbed this photon.

C) What will be energies of all of the photons that could subsequently by emitted by this atom after the photon in part b is absorbed.

D) Determine the wavelength of all the photons found in part c and determine in which part of the electromagnetic spectrum they fall. Determine their approximate color.

E) Sketch the emission spectrum for this element showing only those lines that fall in the visible light range of EM spectrum. Label the lines with their wavelength.

Answer 1

If 17.5eV photon is absorbed by an atom of this substance ,

states was the atom before it absorbed this photon = 3 rd state .

energies of all of the photons that could subsequently by emitted = 30.5 eV , 32.5 eV , 35 eV

=> wavelengths =   (6.63 * 10^-34 * 3 * 10⁸)/(30.5 * 1.6 * 10^-19)

                            = 40.758 nm

=>   wavelengths =   (6.63 * 10^-34 * 3 * 10⁸)/(32.5 * 1.6 * 10^-19)

                            = 38.25 nm

=> wavelengths =   (6.63 * 10^-34 * 3 * 10⁸)/(35 * 1.6 * 10^-19)

                            = 35.51 nm

=>   part of the electromagnetic spectrum they fall = ultraviolet region .