Question

For the hydrogen atom, its energy at ground state is 13.6 eV, at first excited state is 3.4 eV at second excited state is 1.5 eV and at the third excited state is 0.85 eV. i) Give the energy value for the first two states in Joule (J). [1eV =1.6 x 10-19 J] (2 marks) ii) With the aid of schematic diagram, determine the energy of emitted photon when the atom jumps from the first and third excited states to the ground state. (6 marks)

Answer 1

Given: Eg=13.6ev E,=3.4ev E2=1.5 ev Eg=8.85eV 1) Let Eg = 13.681.6x10ls ::Eg = 2.176X1518] 2) E, = 3.4x16xlolos E,= 5,44x10ls PE,=0.544x1olej

The energy of emitted photon when the atom jumps from the first state to ground state is, E-150Eg= 13.4-13-6) :E=10.2 ev similarly cohen atom jumps from the third state to ground state E = 1 E-Egl 10.85-13.61 --5:12.75 e V Ez E₂ 39 19 EL Egg 5,9 Eg