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: 80

Answer

See answers.

Work Step by Step

The 12.3 V potential difference means the electrons have a kinetic energy of 12.3 eV. When they collide with the hydrogen atom, they may transfer this energy. Starting from the ground state, the H atom might have as much as $-13.6eV+12.3eV=-1.3eV$ of energy. The energy level diagram in Figure 27–29 shows us that this corresponds to exciting the H atom up to the n = 3 state. As the atom returns to the ground state, there are 3 possible transitions. Equation 27-15b says that for hydrogen, the energy of a level is $E_n=\frac{-13.6eV}{n^2}$. Combine this with equation 27–10. $E_1$, $E_2$, and $E_3$ are shown in the textbook. Calculate the wavelengths. From n = 3 to n = 2. $$\lambda=\frac{hc}{E_3-E_2}=\frac{1240eV\cdot nm}{-1.51eV-(-3.40eV)}=650nm$$ From n = 3 to n = 1. $$\lambda=\frac{hc}{E_4-E_2}=\frac{1240eV\cdot nm}{-1.51eV-(-13.6eV)}=102nm$$ From n = 2 to n = 1. $$\lambda=\frac{hc}{E_2-E_1}=\frac{1240eV\cdot nm}{-3.40eV-(-13.6eV)}=122nm$$
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.