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 801: 69

Answer

$4.4\times10^{26}\frac{photons}{s}$

Work Step by Step

The microwave oven’s power rating of 720 W is the energy output per second. Divide this power by the energy of one photon to find the number of photons produced per second. Equation 27–4 gives us the energy of one photon. $$E_{photon}=hf=\frac{hc}{\lambda}$$ Number per second = $\frac{P}{E_{photon}}=\frac{P \lambda}{hc}$ $$=\frac{(720W)(0.122m)}{(6.63\times10^{-34}J \cdot s)(3.00\times 10^8 m/s)}$$ $$=4.42\times10^{26}\frac{photons}{s}$$
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