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.

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