PROBSET1. Energy levels of hydrogen : En = n = 5 n = 2Energy of emitted photon: Ei - Ef

E5= = - 0.544 eVE2 = = - 3.4 eV- 0.544 eV (-3.4 eV)2.856 eV x

2. A.When the atom in the n =1 level absorbs a 20 eV photon, the final level of the atom is n = 4. The possible transitions from n = 4 and corresponding photon energies are:n = 4 n = 3:4 eVn = 4 n = 2:10 eVn = 4 n =1:19eVOnce the atom has gone to the n = 3 level, the following transitions can occur:n = 3 n = 2:6 eVn = 3 n =1:15 eVOnce the atom has gone to the n = 2 level, the following transition can occur:n = 2 n =1:9 eVThe possible energies of emitted photons are: 4 eV, 6 eV, 9 eV, 10 eV, 15 eV, and 19 eV.

B.E = hfE = hc/= hc/ E 4.00 eV = = 3.102 x 10-7 m or 310.2 nm 6.00 eV = = 2.068 x 10-7 m or 206.8 nm 9.00 eV = = 1.378666667 x 10-7 m or 137.8666667 nm 10.00 eV = = 1.2408 x 10-7 m or 124.08 nm 15.00 eV = = 8.272 x 10-8 m or 82.72 nm 19.00 eV = = 6.530526316 x 10-8 m or 65.30526316 nm

Legend UV - X-ray

C. There is no energy level 8 eV higher in energy than the ground state, so the photon cannot be absorbed.

3. A. absorbsEnergy of absorbed photon: Ef EiEf = ?Ei = - 6.25 eV

Ef2.311395349 x 10 -19 J x + (-6.25 eV) = Ef Ef = -6.25 eV + 1.444622093 eVEf = - 4.805377907 eV

B. emitsEnergy of emitted photon: Ei - EfEf = ?Ei = -2.68 eVEf Ef732857143 x 10-19 J x + 2.68 eV = Ef 958035714 eV + 2.68 eV = Ef638035714 eV = EfEf = - 638035714 eV