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 802: 90

Answer

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Work Step by Step

a) We sketched the energy level diagram of this unknown element, as you see in the figure below. b) When the atom abosorbs an entry of 5.1 eV, it sends it to the third excited state at $n=4$ which has an energy of -6.4 eV, as we see in the figure below. The eenrgyies of the emitted photons will be as follows: - When the electron moves from $n=4$ to $n=3$. $$E_p=\Delta E_e=E_4-E_3=-6.4-(-6.8)=\color{red}{\bf 0.4}\;\rm eV$$ - When the electron moves from $n=4$ to $n=2$. $$E_p=\Delta E_e=E_4-E_2=-6.4-(-9)=\color{red}{\bf 2.6}\;\rm eV$$ - When the electron moves from $n=4$ to $n=1$. $$E_p=\Delta E_e=E_4-E_1=-6.4-(-11.5)=\color{red}{\bf 5.1}\;\rm eV$$ - When the electron moves from $n=3$ to $n=2$. $$E_p=\Delta E_e=E_3-E_2=-6.8-(-9)=\color{red}{\bf 2.2}\;\rm eV$$ - When the electron moves from $n=3$ to $n=1$. $$E_p=\Delta E_e=E_3-E_1=-6.8-(-11.5)=\color{red}{\bf 4.7}\;\rm eV$$ - When the electron moves from $n=2$ to $n=1$. $$E_p=\Delta E_e=E_2-E_1=-9-(-11.5)=\color{red}{\bf 2.5}\;\rm eV$$
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