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Material Type: Assignment; Professor: Zorba; Class: Optics & Modern Physics; Subject: Physics; University: Whittier College; Term: Unknown 1989;
Typology: Assignments
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Chapter 3: 5 - ) Because is the power per unit area per unit wavelength, while is the power per unit area per unit frequency. In other words, the argument would have been valid if the left sides of these equations were the power per unit area, but they are not. The two formulas are not equal to each other, rather they are related by. 8 - ) (a) (b) (c) # of photons per second per second. 11 - ) The radiation energy per volume per unit wavelength Since we are asked to find the number of visible photons, let us write in terms of total number of photons per volume per wavelength by dividing the by the energy of each photon Thus. If we integrate the right-hand side over between 400- 750 nm for room temperature, we will get , the total number of visible photons per unit volume: . The total number in our room would be photons, which is almost nonexistent!
Thus the fraction is 25 - ) 27 - ) 36 - ) (which is in the UV portion of the EM spectrum). 38 - ) In the Bohr model, the speed of the orbiting electron is given by. Thus the largest speed is The smallest speed is 41 - ) If gravity were the responsible force holding the atom together, we would proceed exactly the same way as in the original case except that we would replace with G , where G (6.67x , the gravitational constant, replaces the electrostatic (Coulomb) constant , and e’s would be changed by the masses of the electron and proton. Thus let us write the original expression we have derived in the class for the electrostatic case and then replace the above mentioned parameters: