Question

1/6 points l Previous Answers AtPChem 11 7AE004a. Calculate the energy of the quantum involved in the excitation of the following systems. (Express the results in joules and kilojoules per mole (a) an electronic oscillation of period 4.8 fs 17 130-19 (b) a molecular vibration of period 48 fs kJ mol-1 (c) a pendulum of period 4.8 s kJ mol-

Answer 1

Given T = 4.8 Fr Now we know that energy of that type system is given by E = hf where h is the Planck�s constant = 6.63 times 10^-34 now f = 1/T implies 1/4.8 times 10^-15 = 2.083 times 10^14 Hz Now E = 6.63 times 10^-34 times 2.083 times 10^14 = 13.8125 times 10^-20 E = 1.38 times 10^-19j now 1 mole = 6.022 times 10^23 Now E = 1.38 times 10^-19 j/6.022 times 10^23 mol implies 2.29 times 10^-43j/mol og E = 2.29 times 10^-43 J/mol times 1kj/10^2 J E = 2.29 times 10^-46 KJ/mol (b) Now T = 48 Fr therefore f = 1/T = 1/48 times 10^-15 implies 2.083 times 10^13 Hz Now E = 9.93 times 10^-34 times 2.083 times 10^13 implies 13.8125 times 10^-21 E = 1.38 times 10^-20 J and E = 1.38 times 10^-20 J/6.022 times 10^23 mol implies 2.29 times 10^-44 J/mol