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: 37

Answer

1.592 MeV $\lambda=7.81\times10^{-13}m$

Work Step by Step

The photon’s energy equals the sum of the total energy of the other 2 particles. $$E=2(KE+mc^2)=2(0.285MeV +0.511MeV)=1.592MeV$$ Now find the wavelength using equation 27–6. $$\lambda=\frac{hc}{E}$$ $$\lambda=\frac{(6.63\times10^{-34}J \cdot s) (3.00\times10^8m/s)}{(1.592\times10^{6}eV)(1.60\times10^{-19}J/eV) }$$ $$\lambda=7.81\times10^{-13}m$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.