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Extra Credit 25

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    195751
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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)

    #60,  #66, #94

    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)

    #16, #39, #52

    Jamal Nasim

    Question 60: 

    The radiant energy from the sun reaches its maximum at a wavelength of about 500.0 nm. What is the approximate temperature of the sun’s surface? 

    Solution: 

    This is a straightforward application of Wien's Displacement Law,

    \[\lambda_{max}=\frac{b}{T} \]

    where \(b\) is Wien's Displacement constant.

    \[b = 2.89\times 10^{-3} \: mK\]

    To solve for temperature, we isolate T. 

    \[T=\frac{b}{\lambda_{max}}\]

    Plug in Wien's Displacement constant and wavelength to find the approximate surface temperature of the sun.

    \[ \begin{eqnarray*} T &=& \frac{2.89 \times 10^{-3} \: mK}{500.0 \times 10^{-9} \: m}\\ &=& 5780 \: K \end{eqnarray*} \]

     

    Question 66: 

    The work function for potassium is 2.26 eV. What is the cutoff frequency when this metal is used as photoelectrode? What is the stopping potential when for the emitted electrons when this photoelectrode is exposed to radiation of frequency 1200 THz?

    Solution: 

    For the photoelectric effect, the following equation relates the kinetic energy of an electron with the frequency of incoming light. 

    \[KE_e = h\nu - \Phi\]

    The cutoff frequency occurs when the photoelectrode is no longer capable of releasing electrons. In other words, the kinetic energy of the electrons is zero. 

    \[KE_e = 0 \]

    Isolate \(\nu\) to find cutoff frequency. 

    \[ \begin{eqnarray*} h\nu &=& \Phi\\ \nu &=& \frac{\Phi}{h}\\ &=& \frac{(2.26 \: eV)\frac{(1.6 \times 10^{-19}\: J}{1 eV})}{6.626\times10^{-34} \: Js}\\ &=& 5.46 \times 10^{14} \: Hz \end{eqnarray*}\]

    To find stopping potential, the following revised equation is used: 

    \[eV_{stop} = h\nu - \Phi \]

    Divide by the fundamental charge \(e\) to find the stopping voltage. 

    \[ \begin{eqnarray*} V_{stop} &=& \frac{h\nu - \Phi}{e}\\ &=& \frac{(6.626\times10^{-34} \: Js)(1200\times 10^{12}\: Hz)(\frac{1 \: eV}{1.6\times10^{-19} \: J}) -2.31 \: eV}{1 \: e}\\ &=& 2.660 \: V \end{eqnarray*}\]

     

    Question 94: 

    When a hydrogen atom is in its ground state, what are the shortest and longest wavelengths of the photons it can absorb without being ionized?

    Solution: 

    Using the Bohr model of the hydrogen atom, the following equation represents the energy transitions between two electronic states: 

    \[E_{photon} = - \frac{e^4m_e}{8\epsilon_0^2h^2} (\frac{1}{n_2^2}-\frac{1}{n_1^2}) \]

    We can simplify the constants by introducing Rydberg energy units \(R_y\).

    \[R_y= \frac{e^4m_e}{8\epsilon_0^2h^2} = 2.18\times 10^{-18} J\]

    To find the wavelength of the shortest and longest absorptions, plug in wavelength in place of energy. 

    \[E_{photon}= \frac{ch}{\lambda}\]

    Plug this definition into the equation for the transitions, and isolate for \(\lambda\): 

    \[ \begin{eqnarray*} \frac{ch}{\lambda} &=& - (\frac{1}{n_2^2}-\frac{1}{n_1^2})R_y\\ \lambda &=& -\frac{hc}{(\frac{1}{n_2^2}-\frac{1}{n_1^2})R_y} \end{eqnarray*} \]

    For absorptions from the ground state, \(n_1 = 1\). Therefore: 

    \[\lambda = \frac{hc}{(1-\frac{1}{n_2^2}) R_y} \]

    The longest wavelength occurs when the denominator is minimized, which occurs at \(n_2 = 2\). 

    \[ \begin{eqnarray*} \lambda &=& \frac{(6.626\times10^{-34} \: Js)(3.00\times10^8 \: m/s)}{(1-\frac{1}{2^2}) (2.18\times10^{-18} \:J)}\\ &=& 1.216\times10^{-17}\:m \end{eqnarray*} \]

    To find the shortest wavelength, the denominator must be maximized, which occurs at \(n_2=\infty). Such a energy level would result in ionization; however, this is a limit, meaning it will suffice in finding the shortest wavelength without ionization. 

    \[ \begin{eqnarray*} \lambda &=& \frac{(6.626\times10^{-34} \: Js)(3.00\times10^8 \: m/s)}{(1-\frac{1}{\infty^2}) (2.18\times10^{-18} \:J)}\\ &=& 9.118\times10^{-8}\:m \end{eqnarray*} \]

     

    Question 16: 

    The metals sodium, iron, and molybdenum have work functions 2.5 eV, 3.9 eV, and 4.2 eV, respectively. Which of these metals will emit photoelectrons when illuminated with 400 nm light?

    Solution: 

    A photon with a wavelength of 400 nm has an energy of \(3.1 \:eV\). This means only sodium will emit photoelectrons when illuminated by this light. 

     

    Question 39: 

    Which type of radiation is most suitable for the observation of diffraction patterns on crystalline solids; radio waves, visible light, or X-rays? Explain.

    Solution: 

    X-rays are most suitable for the observation of diffraction patterns on crystalline solids. Crystalline solids are observed at the atomic level, which is about 0.1 nanometers. As such, the light used must have a wavelength smaller than that, which only x-rays do. 

     

    Question 52: 

    Can the de Broglie wavelength of a particle be known exactly?

    Solution: 

    Yes, the de Broglie wavelength of a particle can be known exactly, at least theoretically. This is because according to the Heisenberg Uncertainty Principle, the product of the uncertainty of the linear momentum with the uncertainty of position is greater than \(\frac{hbar}{2}\). This means that linear momentum can be known exactly, but the position would have infinite uncertainty. Because the de Broglie wavelength is inversely proportional to the linear momentum, it too will be known exactly if the momentum is known exactly. 


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