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Homework Help: Questions related to waves.

  1. Apr 13, 2008 #1
    Hey, I'm having a few problems with the questions below, now I think alot of it is to do with not knowing how to approach the question. Please forgive me if there is not alot of working out, but I may just need to directing in the right direction.

    Question 1
    The average wavelength of light emitted from an incandescent torch bulb with a metal filament is 120nm. Calculate the number of photons emitted by a 20W torch bulb in one hour.

    Answer 1
    Photon energy is proportional to the frequency of the wave.

    [tex]3\times10^8 = f\times 120nm[/tex]

    [tex]\frac{3\times10^8}{120\times10^{-9} = f[/tex]


    I have the frequency now, but how do I get from here to finding how much is emitted by a 20W torch bulb in 1 hour?

    Question 2
    A photon has a momentum given E/c where E is the enerrgy of the photon and c is the speed of light. If the torch bulb emits parallel beam light, then calculate the force on the torch.

    Answer 2
    I have no idea, at all. I am not asking for the answer, but could someone please direct me in the direction of a method of some sort, even if it is only the intial stages.

    Question 3
    Calculate the initial acceleration of the toch if it was in empty space, and it had a mass of 200g.

    Answer 3
    Again, not idea. I know I have the mass, but that is the only value I have. It could be possible that I need values from previous questions above.

    I apologise for th elack of working, but the whole thing has me stumped. I know it is against PF regulations to just dish out the answers, but I am willing to work through it, all I need is a gentle push in the right direction! :bugeye:

    Any help is much appreciated.

    Last edited: Apr 13, 2008
  2. jcsd
  3. Apr 13, 2008 #2


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    for Q#1

    -What is the relation between power(measured in watts: W), energy and time?
    -How is the energy of a photon related to its wavelenght / frequency?

    for Q#2

    - Find the relation between linear momentum and force using the definition of linear momentum and Newtons second law.

    for Q#3

    - Do Q#2 first
    Last edited: Apr 13, 2008
  4. Apr 13, 2008 #3

    A watt is 1 joule of energy per second.

    The energy of a photon is proportional to it's frequency. If E is constant, then an increase in frequency will result in a decrease in wavelength.

    I would be able to convert to W now, but it the convertion to Joules in which I am struggling with.
  5. Apr 13, 2008 #4


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    So how can anyone help you if you are not showing what you did?

    And WHAT are you trying to convert to Joules?

    have you seen this formula:
    [tex] E_{\gamma} = hf = hc/\lambda [/tex]
  6. Apr 13, 2008 #5
    I have shown you everything I know how to do. For question 1, I would have thought I would get an answer in Joules, and then convert to Watts.

    I have not seen [tex] E_{\gamma} = hf = hc/\lambda [/tex] but I have seen [tex]E_{\gamma} = hf[/tex]

    The thing is, I have not used either in school and it is not in the curriculum. If that equation can be used, then I will use it, but I am not sure if there might be an easier way. I will use this one, if you say it will work then.

    [tex] E_{\gamma} = hf = hc/\lambda [/tex]
    [tex] E_{\gamma} = 2.5\times10^{15}h = \frac{3\times10^8h}{120\times10{-9}}[/tex]

    I have looked up Planck's Constant and I will use [tex]6.6\times 10^{-34}[/tex] as the value.

    [tex] E_{\gamma} = 2.5\times10^{15}h = \frac{3\times10^8h}{120\times10{-9}}[/tex]

    [tex] E_{\gamma} = 2.5\times10^{15}\times 6.6\times10^{-34}=\frac{3\times10^8\times6.6\times10^{-34}}{120\times10{-9}[/tex]

    [tex] E_{\gamma} = 1.65\times10{-18} Joules/s[/tex]

    If this is correct then I would multiply my answer by 3600, to get to Hours.
  7. Apr 13, 2008 #6


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    check the units of [tex] E_{\gamma} [/tex].... Joules/s is totaly madness!

    Why not just calculate how much energy the torch bulb emits under 1h, and then evaluate the number of photons with wavelength 120nm that energy corresponds to?
  8. Apr 13, 2008 #7
    I have no idea on the units, planck's constant is in [tex]m^2 kg / s[/tex] How do I convert this to J/s?? I think I will do it this way Malawi, the other way will be explained in class but atleast now, I know another method. I am unsue on how to convert my asnwer to J/s.
  9. Apr 13, 2008 #8


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    But the units of Energy is J, then you can't get an answer with J/s:

    [tex] E_{\gamma} = 1.65\times10{-18} Joules/s[/tex]

    As you wrote.

    This also helps: m^2 kg/s = J*s (from Newtons second law and the fact that 1J = 1N*m)

    The way you do it is wrong, why not do it the correct way which is the one I told you?
    "Calculate the number of photons emitted by a 20W torch bulb in one hour."

    The energy relased by the buld in 1h is 20*3600J, right?

    One photon with wavelenght 120nm has energy hc/lamda = 6.626*10^-34[Js]*3*10^8(m/s) / (120*10-^9(m)) = 1.655*10^-18J (pretty much as you got, but you got wrong units).
  10. Apr 13, 2008 #9
    I follow that now.

    I have a total energy of 72000 Joules

    One photon has an energy of [tex]4.35\times10^-8[/tex]

    Therefor, the total number of photons must be [tex]\frac{72000}{4.35\times10^-8} = 4.35\times10^{12}[tex]

    Thank you for your help here, Malawi. My teacher has not shown us the first equation, and so I do not know how he expected us to do it, other than to do some research.


    I can now find the momentum of the photon as I have both E and C.

    [tex]\frac{1.65\times10^{-18}{3\times10^8} = 5.52\times10^{-27}[/tex]

    Now you mentioned Newton's Second law, F=ma. I know that there is an equation that is closely related to this one.

    EDIT: Thank you for all your time Malawi
    Last edited: Apr 13, 2008
  11. Apr 13, 2008 #10
    I can find the momentum as I have both E and c, so I can do E/c
  12. Apr 13, 2008 #11
    Okay, I am going to also use information from question 1 to answer this question.

    Momentum = Force x Time

    Force = Momentum/Time

    1 Hour = 3600 Seconds
    E/c = [tex]5.5\times106{-17}[/tex]

    [tex]Force= \frac{5.5\times10{-17}}{3600s}[/tex]

    [tex]Force = 1.52 \times10^{-20}N[/tex]

    I don't know if that is any good...
  13. Apr 13, 2008 #12


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    force = time derivative of momentum

    You cant GATHER force, force is instanteous.

    So if the bulb casts away 20W photons in the same direction (we was to assue it was a paralell beam), then how can you relate the power of the bulb to the time derivative of momentum, if momenutm = E/c ?

    btw the energy of one photon is 1.655*10^-18 J
  14. Apr 13, 2008 #13


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    maybe this can help you further:

    Force: [tex] F = \frac{dp}{dt} [/tex] units: N

    Momentum: [tex] p=E/c [/tex] units: m*kg/s

    Power: [tex] P = \frac{dE}{dt} [/tex] units: W = J/s = N*m/s
  15. Apr 13, 2008 #14
    Malawi, I really have no idea at all!!
  16. Apr 13, 2008 #15


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    ok, so you are given power and energy. c is just a constant. can you atleast try to relate the known variables with the equations I just gave you?
    Last edited: Apr 13, 2008
  17. Apr 13, 2008 #16
    Students around the world are confused because they don't follow the right sequence.

    * Convert all givens to SI units.
    * List all formulas.
    * Solve algebraically for the unknown without using any numbers.
    * Sustitute all numbers at once, use your calculator one time.
    * Check units.

    Following that sequence, see how easy it is:

    Givens, converted to SI:
    power: P= 20W
    wavelength: lambda = 120 nm = 120X10^-9 m
    time: t = 1 hr = (1 hr)(60 min / 1 hr)(60 s / 1 min) = 3600 s

    energy and power: E=Pt
    energy of a photon: E=hf=hc/lambda
    energy of n photons: E = nhf = nhc/lambda

    algebraic solution:
    n = .....

    substitute numbers ...... use calculator ....

    check units: dimensionless (correct)
  18. Apr 13, 2008 #17


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    no energy is not = Power times t!!

    P = dE/dt, it is different from the AVERAGE Power: P = E/t
  19. Apr 13, 2008 #18
    Is a time variation mentioned in the given problem?
  20. Apr 13, 2008 #19


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    Is the average force mentioned?

    In problem #2 one needs the correct definitions to solve it.
    One can not just take F = p/t (since that is totaly wrong), that is confusing (as you saw how the OP tried to solved it), using the correct defintions is the safest. As you said: "list all formulas"...

    And aslo: why converting everything to SI units? that is not an a priori thing to do...

    We solved this problem later in the PF chat me and _mayday_
  21. Apr 13, 2008 #20
    My comments addressed only "Question 1" (constant power dissipation).
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