1162_Physics ProblemsTechnical Physics

1162_Physics ProblemsTechnical Physics - Chapter 41 P41.34...

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Chapter 41 503 P41.34 Problem 43 in Chapter 16 helps students to understand how to draw conclusions from an identity. ψ = Axe bx 2 so d dx Ae bx Ae bx bx =− −− 22 2 2 and d dx bxAe bxAe b x e b b x bx bx bx 2 2 23 244 6 4 2 ψψ + + . Substituting into Equation 41.13, −+ = F H G I K J + F H G I K J 64 2 2 2 bb x mE m x ω == . For this to be true as an identity, it must be true for all values of x . So we must have both −= 6 2 2 b mE = and 4 2 2 b m = F H G I K J = . (a) Therefore b m = 2 = (b) and E b m 33 2 2 = = . (c) The wave function is that of the first excited state . P41.35 The longest wavelength corresponds to minimum photon energy, which must be equal to the spacing between energy levels of the oscillator: hc k m λ so λπ π × × F H G I K J = 3 0 0 1 0 911 10 600 8 31 12 c m k . . ms kg 8.99 N m nm ej . P41.36 (a) With =
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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