Extra Credit 29

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https://phys.libretexts.org/Bookshel...ves_(Exercise)

Solve (i.e., write reasonable step-by-step tutorials for the solutions) of these exercises from the text)

#64, #97, #108

Answer (i.e., write the answer to these problems, but no need to show work. You can if you prefer)

https://phys.libretexts.org/Bookshel...ves_(Exercise)

#65, #98, #109

Problem 64:

The longest wavelength of radiation is when \[\frac{hc}{\lambda } = \Phi\]

\[\lambda = \frac{hc}{\Phi} = \frac{(6.626\times10^{-34}Js) (3\times10^{8}\frac{m}{s})}{2.24eV\times\frac{1.6\times10^{-19}J}{1eV}} =5.55\times10^{-7}m =555 nm\]

It is in the visible range.

Problem 97:

\[a)\hspace{1mm} Angular \hspace{1mm} momentum = n\hbar = 2(1.05\times10^{-34}Js) =2.1\times10^{-34}Js \\ b)\hspace{1mm} r= a_{0}n^{2}=(5.29\times10^{-11}m)(2^2) = 2.12\times10^{-10}m \\ Kinetic \hspace{1mm} energy = \frac{1}{2}mv^2 = \frac{p^2}{2m} =\frac{n^2\hbar^2}{2mr^2} = \frac{(2.1\times10^{-34}Js)^2}{2(9.109\times10^{-31}kg)(2.12\times10^{-10}m)^2} =5.39\times10^{-19}J \\ c) \hspace{1mm} Potential \hspace{1mm} energy = -\frac{e^2}{4\pi\varepsilon_0r} = - \frac{(1.602\times10^{-19}c)^2}{(1.113\times10^{-10}\frac{c^2}{Jm})(2.12\times10^{-10}m)} =-1.09 \times 10^{-18}J \\ d) \hspace{1mm} Total \hspace{1mm}energy = KE +PE =5.39\times10^{-19}J - 1.09\times10^{-18}J =-5.51\times10^{-19}J\]

Problem 108:

\[KE = eV = \frac{1}{2}mv^2=\frac{p^2}{2m} \\ p = \sqrt{2meV} \\ de \hspace{1mm} Broglie \hspace{1mm} wavelength \hspace{1mm} is \hspace{1mm} found \hspace{1mm} by: \\ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}} = \frac{6.626\times10^{-34}Js}{\sqrt{2(9.1\times10^{-31}kg)(1.6\times10^{-19}C)(20\times10^3V)}} = 8.7 pm\]

Problem 65:

\[ \Phi= \frac{hc}{\lambda} = \frac{(6.626\times10^{-34}Js) (3\times10^{8}\frac{m}{s})}{337\times10^{-9}m} =5.9 \times10^{-19} J\]

Problem 98:

\[ \frac{1}{\lambda}=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}] \\ \frac{1}{\lambda}=R[\frac{1}{n_1^2} - 0] \\ \lambda = \frac{n_1^2}{R}=\frac{16}{1.097\times10^7} \\ E = \frac{hc}{\lambda} =\frac{hcR}{n_1^2} = \frac{(6.626\times10^{-34}Js)(3\times10^8\frac{m}{s})(1.097\times10^7m)}{4^2} =1.36\times10^{-19}J \]

Problem 109:

For KE = 2.0MeV:

\[ \lambda = \frac{6.626\times10^{-34}Js}{\sqrt{2(1.672\times10^{-27}kg)(1.6\times10^{-19}C)(2.0\times10^6V)}} = 0.020 pm\]

For KE = 10.0MeV:

\[ \lambda = \frac{6.626\times10^{-34}Js}{\sqrt{2(1.672\times10^{-27}kg)(1.6\times10^{-19}C)(10.0\times10^6V)}} = 9.06\times10^{-3}pm\]