Physical Chemistry II – Exam 1 SOLUTIONS

3.) (17 points) An electron in a one-dimensional box of width 6.0 Å undergoes atransition from an initial state with n=2 to some final state. The wavelength of lightabsorbed in this transition was determined to be 56.5 nm. Calculate the quantumnumber of the final state. [1 Å = 10 –10 m; 1 nm = 10 –9 m]4The particle in a box energies are given byE n = n2 h 28mL 2 .For a transition from n=2 to some final state n, the energy difference€ΔE = E n − E 2€= n2 h 28mL 2 − 22 h 28mL 2( )ΔE = n 2 − 2 2h 28mL 2 .ΔE isSolving for the unknown quantum number n yields,€n 2 − 2 2n 2= 8mL2 ΔEh 2= 4 + 8mL2 ΔEh 2 .€A photon with an energy corresponding toE photon = ΔE = hν. Since, € for light, λν = c , we can substitute ν = λ , and calculate thecenergy difference€ΔE ,€ΔE would have a frequency given by€ΔE = hc€λ( 6.62607 ×10 −34 Js) 2.99793 ×10 8 m/s=56.5 ×10 −9 mΔE = 3.5158 ×10 −18 J .( )€