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 - Problems - Page 800: 61

Answer

-0.544 eV

Work Step by Step

Use the angular momentum to find the quantum number n, from equation 27-11. $$L=n\frac{h}{2\pi}$$ $$n=\frac{2\pi L}{h}$$ $$n=\frac{2\pi (5.273\times10^{-34}kg \cdot m^2/s)}{6.626\times10^{-34}J \cdot s}=5$$ Now find the energy using equation 27–15b with Z = 1 and n = 5. $$E_{n}=\frac{-13.6eV(Z^2)}{n^2}$$ $$E_{n}=\frac{-13.6eV(1^2)}{5^2}=-0.544eV$$
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