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 - General Problems - Page 802: 92

Answer

a) $0.4\;\rm V$ b) $3.75\times10^5\;\rm m/s$ c) $1.94\;\rm nm$

Work Step by Step

a) We know that the energy of the electron is given by $$E=W_0+KE$$ Thus, its kinetic energy is given by $$KW=E-W_0= eV$$ So that the needed voltage is given by $$V=\dfrac{E-W_0}{e}$$ Recalling that $E=hf=\dfrac{hc}{\lambda}$; $$V=\dfrac{\dfrac{hc}{\lambda}-W_0}{e}$$ Plugging the known; $$V=\dfrac{\dfrac{6.626\times10^{-34}\times 3\times10^8}{ 464\times10^{-9}}-(2.28\times1.6\times10^{-19})}{1.6\times10^{-19}}$$ $$V=0.397\;\rm V\approx \color{red}{\bf 0.4}\;\rm V$$ --------------------- b) First, we need to find the electron speed by using the normal kinetic energy law and then compare it to the speed of light ($0.1C$), if it was too small to be compared with it so be it, if not, we will need to use the relativistic kinetic energy law. $$KE=\frac{1}{2}m_ev^2$$ Solving for $v$; $$v=\sqrt{\dfrac{2KE}{m_e}}=\sqrt{\dfrac{2eV}{m_e}}$$ Plugging the known; $$v=\sqrt{\dfrac{2\times1.6\times10^{-19}\times0.4 }{9.11\times 10^{-31}}}=\color{red}{\bf 3.75\times10^5}\;\rm m/s$$ which is about $0.00125c$, so our result is acceptable. --------------------- c) We know that the wavelength is given by $$\lambda=\dfrac{h}{p}=\dfrac{h}{m_ev}$$ Plugging the known; $$\lambda= \dfrac{6.626\times10^{-34}}{9.11\times10^{-31}\times 3.75\times10^5}=\color{red}{\bf1.94}\;\rm nm$$
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