Extra Credit 28
- Page ID
- 195754
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Solve (i.e., write reasonable step-by-step tutorials for the solutions) of these exercises from the text)
#63, #96, #107
#63
We start with the equation
\[KE = h\nu - \phi \nonumber\]
Since we want the largest wavelength, we assume that the KE is the smallest possible with is 0 in this case giving us
\[h\nu = \phi \nonumber\]
Frequency is speed of light divided by wavelength so we substitute frequency with wavelength.
\[h* (\frac{c}{\lambda}) = \phi \nonumber\]
Given at the binding energy of silver is 4.73 eV converted to 7.578295e-19 Joules.We solve for lambda to get
\[\lambda =262 nm \nonumber\]
This is outside the range for visible light which is 380 to 700 nm
#96
\[\Delta E = -R_{H} (\frac{1}{n_{f}^2}-\frac{1}{n_{i}^2}) \nonumber\]
where RH is 2.178E-18 J and the n are the final and initial energy states respectively
We will also be using
\[\Delta E = \frac{hc}{\lambda}\nonumber\]
so
\[\lambda = \frac{hc}{\Delta E}\nonumber\]
\[n_{f} = \infty \nonumber\]
\[n_{i} = 1 \nonumber\]
\[\Delta E = -R_{H}(\frac{1}{\infty^2}-\frac{1}{1^2}) = -1.06 E-19 J \nonumber\]
\[\Delta E = 2.178 E-18 J \nonumber\]
\[\lambda = \frac{hc}{\Delta E}\nonumber\]
\[\lambda = 9.10E-8m\nonumber\]
#107
We use the formula for de Broglie wavelength
\[\lambda = \frac{h}{p} = \frac{h}{mv} \nonumber\]
plug in the mass of an electron for m an dtehh speed for v where h is the plank constant
\[\lambda = \frac{h}{(9.10938356*10^{-3} )( 5.0*10^{6})} \nonumber\]
\[\lambda = 1.4556*10^{-10}m \nonumber\]
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)
#64, #97, #108
#64
We start with the equation
\[KE = h\nu - \phi \nonumber\]
Since we want the largest wavelength, we assume that the KE is the smallest possible with is 0 in this case giving us
\[h\nu = \phi \nonumber\]
Frequency is speed of light divided by wavelength so we substitute frequency with wavelength.
\[h* (\frac{c}{\lambda}) = \phi \nonumber\]
Given at the binding energy of potassium is 2.24 eV converted to 3.58888e-19 Joules.We solve for lambda to get
\[\lambda =5.54 *10^{-7} m \nonumber\]
#97
For an electron in a hydrogen atom in the n=2 state, compute:
(a) the angular momentum;
L = 2E-33 Js
(b) the kinetic energy;
3.4 eV
(c) the potential energy; and
-6.8 eV
(d) the total energy
-3.4 eV
#108
What is the de Broglie wavelength of an electron that is accelerated from rest through a potential difference of 20 kV?
0.274 nm