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Example text

Thus. 59. 61) Since the terms on the left are functions only of r and the terms on the right are functions only of 0. both sides of this equation must be equal to the same constant, 0 say. Thus we obtain two ordinary differential equations, one for R(r) and one for 6(0), namely sin 0

Since the sine function vanishes at all multiples of TT. we see that k must be chosen as 717T fc=— , ra = l , 2 , 3 , . . 48) a These discrete or quantized values of k are denoted by kn. Thus, the only allowed solutions of the Schrodinger wave equation are the functions T17TX ^n(x) = Asm - , n = l , 2 , 3 , . . 49) a where A is any non-zero constant. Moreover, the particle's energy, E = /i 2 /c/(87T 2 m), can have only discrete values given by, L, L2 2 2 n = 1,2,3,.... 50) These discrete values En are sometimes called the eigenvalues of Eq.

18) where A is the amount of energy (the so-called work function) required to free an electron from the metal. Thus if hv < A, no photoelectrons are produced. Xcurrent \ collector metal surface per unit time and, thus. Wmeter \/ the rate of photoelectron emission. X' photoelectron . 3. A schematic illustration of stein's prediction of the maximum enthe experimental arrangement used to verergy of a photoelectron, Eq. 18), ify photoelectric effect. ^ verified quajltitatively using the experiment shown schematically in Fig.

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Eval of Prompt Fission Gamma Rays for Simulating Nuclear Safeguard Measurements


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