23) What is the ratio of the de Broglie wavelength of the electron in the n = 4

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Explanation / Answer

We know that En = -13.6 eV / n^2 ==> h c / lambda = -13.6 eV /n^2 ==> lambda = n^2 hc / 13.6 eV for n = 4, we have lambda 4 = 16 * 6.6 x 10^-34 * 3 x 10^8 / 13.6 x 1.602 x 10^-19 = 14.5406 x 10^-7 m = 1454.06 x 10^-9 m = 1454.06 nm for n = 1, we have lambda 1 = 1 * 6.6 x 10^-34 * 3 x 10^8 / 13.6 x 1.602 x 10^-19 = 0.9087 x 10^-7 m = 90.87 x 10^-9 m = 90.87 nm Therefore lambda 4 / lambda 1 = 1454.06 nm / 90.87 nm = 16.00

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