Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 27 - Early Quantum Theory and Models of the Atom - Search and Learn - Page 802: 3

Answer

a. 2.0 nm. b. It would not be possible to construct the 2 slits.

Work Step by Step

a. The KE of 12 eV is much less than the rest energy of an electron. We may use nonrelativistic relationships. Calculate the wavelength using equation 27–8. $$KE=\frac{p^2}{2m}$$ $$p=\sqrt{2m(KE)}$$ $$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2m(KE)}}$$ $$\lambda=\frac{6.626\times10^{-34}J \cdot s }{\sqrt{2(9.11\times10^{-31}kg)(12eV)(1.60\times10^{-19}J/eV)}}$$ $$=3.543\times10^{-10}m $$ Now use equation 24–2a for 2-slit interference. $$m\lambda=d sin \theta$$ Solve for the slit separation. $$d=\frac{m \lambda}{sin \theta}=\frac{(1)( 3.543\times10^{-10}m)}{sin 10^{\circ}}$$ $$d=2.0\times10^{-9}m$$ b. The required separation distance between the 2 slits is about the same size as an atom, so constructing this experiment would be impossible.
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