Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 30 - Quantum Physics - Problems and Conceptual Exercises - Page 1076: 65

Answer

(a) $1.58\frac{Km}{s}$ (b) $62.4^{\circ}$

Work Step by Step

(a) We know that $\lambda=\frac{h}{p}=\frac{h}{mv}$ This can be rearranged as: $v=\frac{h}{m\lambda}$ We plug in the known values to obtain: $v=\frac{6.63\times 10^{-34}}{1.67\times 10^{-27}\times 0.250\times 10^{-9} }$ $v=1.58\times 10^3\frac{m}{s}=1.58\frac{Km}{s}$ (b) We can find the angle as $2dsin\theta=2\lambda $ $\implies 2\times 0.282\times 10^{-9}\times sin\theta=2\times 0.250\times 10^{-9}$ $\implies sin\theta=0.886$ $\theta=62.4^{\circ}$
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