Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 40 - All About Atoms - Problems - Page 1251: 74

Answer

$\hbar = 1.06\times 10^{-34}~J\cdot s = 6.59\times 10^{-16}~eV \cdot s$

Work Step by Step

$\hbar = \frac{h}{2\pi}$ $\hbar = \frac{6.626\times 10^{-34}~J\cdot s}{2\pi}$ $\hbar = 1.06\times 10^{-34}~J\cdot s$ $\hbar = (1.06\times 10^{-34}~J\cdot s)(\frac{1~eV}{1.6\times 10^{-19}~J})$ $\hbar = 6.59\times 10^{-16}~eV\cdot s$
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