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Extra Credit 20

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

    #55,  #61, #89

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

    #11, #34, #47


    Problem 55

    A 200-W heater emits a 1.5-µm radiation.


    A) What value of the energy quantum does it emit?

    Wavelength can be directly correlated to energy through

    \[ E=hf=\frac{hc}{\lambda} \] 

    plugging in the value for lambda we get

     \[ \frac{h*c}{1.5*10^{-6}}=1.325E-19 J \] 



    B) Assuming that the specific heat of a 4.0-kg body is 0.83 \( \frac{kcal}{kg⋅K} \), how many of these photons must be absorbed by the body to increase its temperature by 2 K?

     Let's first start by finding out how much the 4kg body needs to absorb. We will need  \( 1Kcal=4184J \) to solve this.

    \[ 4.0 kg*2K*0.83\frac{Kcal}{kg*K}*4184\frac{J}{Kcal}=27781.8 J \]

    Since we know each photon is 1.325E-19 J we can just divide the energy by the amount of energy per photon

    \[ \frac{27781.8 J }{1.325E-19 \frac{J}{electron}}=2.0967E23 \; photons  \] 


    C) How long does the heating process in (b) take, assuming that all radiation emitted by the heater gets absorbed by the body?


    Watts are just joules per second. Since we have the total amount of energy, it's just a simple division.

    \[ \frac{27781.8 J}{200\frac{J}{s}}=138.91s \]


    Problem 61

    A photon has energy of 20 keV. What are its frequency and wavelength?


    We know that the photon has energy of 20KeV, and that the energy, wavelength, and frequency are related by this equation

    \[ E_{photon}=\frac{1240ev*nm}{\lambda} \]

    So if we plug in numbers, 

    \[20,000=\frac{1240ev*nm}{\lambda} \]
    Solve for Lambda

    \( \lambda=0.062 nm \)

    Wave length and frequency are related through the speed of light.

    \[ f=\frac{c}{\lambda} \]
    \[ f=\frac{3E8\frac{m}{s}}{6.2E-11m}=4.8387E18 Hz \]





    Problem 89

    Calculate the wavelength of the first line in the Lyman series and show that this line lies in the ultraviolet part of the spectrum.

    Lyman series is from any arbitrary energy level n to energy level 1.
    UV Spectrum is from 100-400nm

    Lyman series is as follows:

    \[ \frac{1}{\lambda}=R(\frac{1}{1} - \frac{1}{n^2}) \] 

    where \( R=1.0967E7 \)

    The first line in the Lyman series is \( n =2 \)

    \[ \frac{1}{\lambda}=R(\frac{1}{1}-\frac{1}{4})=8.225E6=\frac{1}{\lambda} \]

    Thus we get that \( \lambda=121 nm \)

    Since \( 100nm \leq 121nm \leq 400nm \) , we can confirm that the first line in the Lyman series lies in the ultraviolet part of the spectrum.







    Problem 11


    Explain how you can determine the work function from a plot of the stopping potential versus the frequency of the incident radiation in a

    photoelectric effect experiment. Can you determine the value of Planck’s constant from this plot?

    Since \(ev_{stop}=hv-\Phi \) , you are able to determine \( \Phi \) if you know the wavelength and \( ev_{stop} \). However, unless you already know every other variable in this equation, you cannot calculate the constant for h.

    Problem 34

    Discuss why the allowed energies of the hydrogen atom are negative.

    Since you need to add energy to "release" the electron from the proton, the total energy generated is negative.

    Problem 47

    Discuss: How does the interference of water waves differ from the interference of electrons? How are they analogous?

    The interference of water is physical, with the amplitude of the crests and troughs. For electrons, it is a probability distribution, and so there is no corporal wave. The water wave also requires a medium to travel, whereas the electron wavefunction does not. Both waves interfere constructively and destructively and produce an interference pattern.







    Extra Credit 20 is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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