Electronic Structure of Atoms

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

            n_i = 5          n_f = 1

            E = -2.18 x 10^-18 J * (1/n_f² - 1/n_i²)

            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) n_i = 4          n_f = 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) n_i = 4          n_f = 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