Extra Credit 18

Questions

Chapter 7

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

https://phys.libretexts.org/Bookshel...cs_(Exercises)

#55, #68, #85

Chapter 6

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)

#9, #17, #32

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Solutions

#9  (Section 6.3 Photoelectric Effect) 

Question: For the same monochromatic light source, would the photoelectric effect occur for all metals?

The photoelectric effect would not occur for all metals if they were exposed to the same monochromatic light source. This is due to the fact that each metal has its own cut-off frequency, \(f_{c}\), and below this frequency, the photoelectric effect will not occur. A metal's cut-off frequency is dependent on its work function and each metal has a specific work function. 

\[ f_{c}=\frac{\phi}{h} \] See equation 6.3.5 

 

#17 (Section 6.4 The Compton Effect & Section 6.3 Photoelectric Effect) 

Question:  Discuss any similarities and differences between the photoelectric and the Compton effects.

 

                Figure 1: Comparison: Compton vs.Photoelectric was created in Canva to visually display the main points of each effect. The Compton Effect is on the left, and the Photoelectric Effect is on the right. Their similarities are in the center. 

Another main difference in the effects was how they were experimentally conducted. For the photoelectric effect, there was a photoelectrode involved and emitted photoelectron as electrons traveled from the anode (target metal) to the cathode in the evacuated glass tube. This created the photocurrent. For the Compton Effect, an x-ray source was shot at a target through slights then a detector would measure the intensity of tradition scattered at \( \theta \) with respect to the incident wavelength, \( \lambda \)'. The difference in the incident and scattered wavelength became known as the Compton Shift,\( \Delta \lambda \). 

#32 (Section 6.5 Bohr's Model of the Hydrogen Atom) 

Question:  How is the energy conserved when an atom makes a transition from a higher to a lower energy state?

Energy is conserved when an atom makes a transition from a higher to lower energy state by emitting a photon with energy equivalent to the difference of the states. For example, in Hydrogen, this energy can be calculated by \( E=(13.6 \mathrm{eV})\left[\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right] \) or by the Rydberg formula, \( \frac{1}{\lambda}=R_{H}\left(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right) \) where \(R_{H} \) is \( 1.097 \cdot 10^{7} \mathrm{~m}^{-1} \). 

Figure 2: Above shows the energy spectrum of the hydrogen atom(left) where the arrows show transitions from a higher energy state to n=1, the ground state. As these energy transitions occur, a photon is emitted as shown on the right.   

#55 (Section 7.4 The Quantum Particle in a Box) 

Question:  Suppose an electron confined to a box emits photons. The longest wavelength that is registered is 500.0 nm. What is the width of the box?

\( E_{n}=n^{2} \frac{\pi^{2} \hbar}{2 m L^{2}} \)

\( \Delta E_{n+1, n}=E_{n+1}-E_{n} \)

\( E=\frac{h c}{\lambda} \)

\( \Delta E_{21}=4 E_{2}-E_{1}=3 E_{1} \)

\( \Delta E_{21}=\frac{3 \pi^{2} \hbar^{2}}{2 m L^{2}} \)

\( \Delta E_{21}=\Delta E=\frac{h c}{\lambda}=\frac{3 \pi^{2} \hbar^{2}}{2 m L^{2}} \) we can then isolate for L

\( L^{2}=\frac{3 \pi^{2} \hbar^{2} \lambda}{2 m h c} \)

\( L=\sqrt{\left(3 \frac{\pi^{2} \hbar^{2} \lambda}{2 m h c}\right)} \) where \( m=9.106 \cdot 10^{-31} \mathrm{~kg} \), \( c=3 \cdot 10^{8} \mathrm{~m} / \mathrm{s} \),  \( \hbar=\frac{h}{2 \pi}=1.055 \cdot 10 \mathrm{ J \cdot s} \), and \( h=6.626 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s} \)

Plug in the values we know, \( \lambda=500 \mathrm{~nm}=500 \cdot 10^{-9} \mathrm{~m} \)

\( L=\sqrt{\frac{3 \pi^{2}\left(1.055 \cdot 10^{-34}\right)^{2}\left(500 \cdot 10^{-9}\right)}{2\left(9.106 \cdot 10^{-31}\right)\left(6.626 \cdot 10^{-34}\right)\left(3.0 \cdot 10^{8}\right)}} \)

\( L=\sqrt{\frac{1.64776747 \cdot 10^{-73}}{3.6201818136 \cdot 10^{-55}}} \)

\( L=\sqrt{4.5516 \cdot 10^{-19}} \)

\( L=6.74656639 \cdot 10^{-10} \mathrm{~m} \)

\( 1 Å =1.10^{-10} \mathrm{~m} \)

\( L=6.74 Å \)

 

#68 (Section 7.5 The Quantum Harmonic Oscillator) 

Question:  A mass of 0.250 kg oscillates on a spring with the force constant 110 N/m. Calculate the ground energy level and the separation between the adjacent energy levels. Express the results in joules and in electron-volts. Are quantum effects important?

For the ground energy level:

\( E_{n}=\left(\frac{2 n+1}{2}\right) \hbar \omega \)

\( E_{0}=\frac{\hbar \omega}{2} \)

\( \omega=\sqrt{\frac{k}{m}} \)

\( \omega=\sqrt{\frac{110}{0.25}} \)

\( E_{0}=\frac{1.055 \cdot 10^{-34}}{2}\left(\sqrt{\frac{110}{0.25}}\right) \)

\( E_{0}=\frac{1.055 \cdot 10^{-34} \cdot 20.976}{2} \)
\( E_{0}=1.10649333 \cdot 10^{-33} \mathrm{~J} \)
\( E_{0}=1.11 \cdot 10^{-33} \mathrm{~J} \)

\( 1 \mathrm{eV}=1.6 \cdot 10^{-19} \mathrm{~J} \)
\( E_{0}=6.91558 \cdot 10^{-15} \mathrm{eV} \)
\( E_{0}=6.92 \cdot 10^{-15} \mathrm{eV} \)

For the adjacent energy levels:

\( \Delta E=\hbar \omega=2 \cdot E_{0} \)
\( \Delta E=2 \cdot 6.92 \cdot 10^{-15} \mathrm{eV} \)
\( \Delta E=1.38 .10^{-14} \mathrm{eV} \)

Quantum effects are not important since the values of the ground and adjacent are small. 

#85 (Section 7 Additional Problems) 

Question:  Atoms in a crystal lattice vibrate in simple harmonic motion. Assuming a lattice atom has a mass of 9.4×10−26, what is the force constant of the lattice if a lattice atom makes a transition from the ground state to the first excited state when it absorbs a 525-µm photon?

\( \begin{aligned} E_{0} &=\frac{\hbar}{2} \sqrt{\frac{k}{m}} \\ E_{1} &=\frac{3 \hbar}{2} \sqrt{\frac{k}{m}} \\ \Delta E &=E_{1}-E_{0}=\frac{h C}{\lambda} \\ \hbar \sqrt{\frac{k}{m}} &=\frac{h c}{\lambda} \end{aligned} \)

isolate for k, the force constant

\( \hbar=\frac{h}{2 \pi} \)
\( \frac{h}{2 \pi} \sqrt{\frac{k}{m}}=\frac{h c}{h} \)
\( \sqrt{\frac{k}{m}}=\frac{2 \pi c h}{\lambda h} \)
\( \frac{k}{m}=\frac{4 \pi^{2} c^{2}}{\lambda^{2}} \)

\( k=\frac{4 \pi^{2} c^{2} m}{\lambda^{2}} \)

\( m=9.4 \cdot 10^{-26} \mathrm{~kg} \)
\( \lambda=525 \mu \mathrm{m}=525 \cdot 10^{-6} \mathrm{~m} \)

\( k=\frac{4 \pi^{2}\left(3.0 \cdot 10^{8}\right)^{2}\left(9.4 \cdot 10^{-26}\right)}{\left(525 \cdot 10^{-6}\right)^{2}} \)

\( k=\frac{3.33987 \cdot 10^{-7}}{2.75625 \cdot 10^{-7}} \)
\( k=1.211745716 \)
\( k=1.21 \mathrm{~N} / \mathrm{m} \)