Electronic Structure of Atoms

 

28) (a) h = 6.63 x 10-34 J•s          c = 3.00 x 108 m/s

            ni = 5          nf = 1

            E = -2.18 x 10-18 J * (1/nf2 - 1/ni2)

            E = -2.18 x 10-18 J * (1/12 - 1/52)

            E = -2.09 x 10-18 J and because E is negative, energy is emitted

            E = hf

            f = 2.09 x 10-18 J/(6.63 x 10-34 J•s) = 3.16 x 1015 /s

            c = f * wlength

            wlength = 3.00 x 108 m/s/(3.16 x 1015 /s) = 9.50 x 10-8 m

 

      (b) ni = 4          nf = 2

            E = -2.18 x 10-18 J * (1/22 - 1/42)

            E = -4.09 x 10-19 J and because E is negative, energy is emitted

            E = hf

            f = 4.09 x 10-19 J/(6.63 x 10-34 J•s) = 6.17 x 1014 /s

            c = f * wlength

            wlength = 3.00 x 108 m/s/(6.17 x 1014 /s) = 4.87 x 10-7 m

            Because this wavelength or frequency falls in the visible portion of the electromagnetic spectrum, visible light is emitted.

 

      (c) ni = 4          nf = 6

            E = -2.18 x 10-18 J * (1/62 - 1/42)

            E = 7.57 x 10-20 J and because E is positive, energy is absorbed

            E = hf

            f = 7.57 x 10-20 J/(6.63 x 10-34 J•s) = 1.14 x 1014 /s

            c = f * wlength

            wlength = 3.00 x 108 m/s/(1.14 x 1014 /s) = 2.63 x 10-6 m

           

            

          

           

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