What is energy? What is work? List some examples of each.
Solution: Here, we are going to calculate the wavelength of the given radiation. Also, we are going to identify the appliance emitting the radiation. Step1: The distance between two adjacent crests or troughs is called wavelength. It is denoted by the o o -10 greek letter lambda() and is generally expressed in terms of Angstrom units(A ) [1 A = 10 m] On the other hand, the number of waves which pass through a given point in one second is known as the frequency. It is denoted by the greek letter nu() and is generally expressed in terms of Hertz(Hz) [1 Hz = 1 sec ] -1 If c is the velocity of light, the the relation between wavelength, frequency and velocity of light is given by: c = x ------(1) Step2: a) Given, frequency of the radiation = 2450 MHz 6 6 = 2450 x 10 Hz. [1 MHz = 10 Hz] = 2.450 x 10 sec . -1 [1 Hz = 1 sec ] -1 Speed of light (c) = 2.998 x 10 m/sec8 Therefore, from equation (1), wavelength() = c/ 8 9 -1 Substituting the values, we get, = (2.998 x 10 m/sec) / (2.450 x 10 sec ) = 1.22 x 10 m.1 = 0.122 m. Thus, wavelength of the radiation is 0.122 m. Step3: b) No, the radiation is not visible to human eye. A typical human eye will respond to -7 -7 wavelengths from about 390 to 700 nm, i.e., (3.90 x 10 to 7.0 x 10 ) m. The wavelength of the radiation produced by the kitchen appliance is 0.122 m. Since, this wavelength does not fall in the visible region of the electromagnetic radiation, therefore, it will not be visible to human eye. Step4: c) Let us calculate the energy of the photon emitted by the appliance and also the photon of the visible light. 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 ---(2), where h is a universal constant called Planck’s constant. is the frequency and E is the energy of a photon. Or, E = hc/ , where, is the wavelength of light and c is the velocity. Step5: 9 -1 Now, frequency of the emitted radiation by the appliance = 2.450 x 10 sec . h = 6.626 x 10 J.s -34 Substituting the values in equation (2), we get, E = (6.626 x 10 J.s) x (2.450 x 10 sec ) 9 -1 = 16.23 x 10 J -25 -24 = 1.623 x 10 J Again, consider wavelength of the photon of visible range = 3.90 x 10 m -7 Using equation (1), we get, frequency() = c/ 8 -7 Substituting the values, we get, = (2.998 x 10 m/sec) / (3.90 x 10 m) = 0.769 x 10 sec . -1 Substituting the values in equation (2), we get, -34 15 -1 E = (6.626 x 10 J.s) x ( 0.769 x 10 sec .) = 5.095 x 10 J -19 Thus, photons of the radiation emitted by the appliance is lower than the photon of visible light. Step6: d) The wavelength of the radiation emitted by the appliance is 0.1 m which falls in the microwave region of the electromagnetic spectrum. Thus, the appliance is likely to be a microwave oven. ---------------------