Problem 28E

What is the standard enthalpy of formation for a compound? For a pure element in its standard state?

Solution: Here, we are going to calculate and compare the energy of the photons emitted by the given two radio stations. Step1: According to Planck’s quantum theory of radiation, the amount of energy associated with a quantum(in case of light, the quantum of energy is often called photon) of radiation is proportional to the frequency of light, E = h ---(1), where h is a universal constant called Planck’s constant. is the frequency and E is the energy of a photon. Step2: Given, frequency of the photon = 1010 KHz = 1010 x 10 s 3 -1 [1 KHz = 1000 Hz = 1000 s ] -1 = 1.010 x 10 s 6 -1 -34 Planck's constant, h = 6.626 x 10 J.s Substituting the values in equation (1), we get, E = (6.626 x 10 J.s) x (1.010 x 10 s ) 6 -1 -28 = 6.69 x 10 J Thus, energy of one photon emitted by AM radio station is 6.69 x 10 J. -28 Step3: Given, frequency of the photon = 98.3 MHz = 98.3 x 10 s 6 -1 [1 MHz = 10 Hz = 10 s ] 6 -1 7 -1 = 9.83 x 10 s Planck's constant, h = 6.626 x 10 J.s -34 Substituting the values in equation (1), we get, -34 7 -1 E = (6.626 x 10 J.s) x (9.83 x 10 s ) = 65.13 x 10 J -27 = 6.513 x 10 J -26 -26 Thus, energy of one photon emitted by FM radio station is 6.513 x 10 J. Step4: From, equation (1), it is seen that energy of a photon is directly proportional to its frequency, i.e., greater the frequency, greater will be the energy. Thus, the energy of photon emitted by FM radio station is greater than the photon emitted by the AM radio station. ------------------