1162_Physics ProblemsTechnical Physics

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

This preview shows page 1. Sign up to view the full content.

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 =
This is the end of the preview. Sign up to access the rest of the document.

## 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 .

Ask a homework question - tutors are online