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28 August, 12:49

1 Calculate the size of the quantum involved in the excitation of (a) an electronic motion of frequency 1.0 * 1015 Hz, (b) a molecular vibration of period 20 fs, and (c) a pendulum of period 0.50 s. Express the results in joules and in kilojoules per mole.

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  1. 28 August, 16:45
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    Answer: a) E = 6.63x10^-19J

    E = 3.97*10^2KJ/mol

    b) E = 3.31*10^-19J

    E = 18.8*10^4 KJ/mol

    C) E = 1.32*10^-33J

    E = 8.01*10^-10KJ/mol

    Explanation:

    a) E = h * f

    h = planks constant = 6.626*10^-34

    E = (6.626*10^-34) * (1.0*10^15)

    E=6.63*10^-19J

    1mole = 6.02*10^23

    E = (6.63*10^-19) * (6.02*10^23)

    E=3.97*10^2KJ/mol

    b) E = (6.626*10^-34) / (1.0*10^15)

    E=3.13*10^-19J

    E = 3.13*10^-19) * (6.02*10^23)

    E = 18.8*10^3KJ/MOL

    c) E = (6.626*10^-34) / 0.5

    E = 1.33*10^-33J

    E = (1.33*10^-33) * (6.02*10^23)

    E = 8.01*10^-10KJ/mol
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