Questions related to waves.

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

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

    [tex]f=2.5\times10^{15}[/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.

    _Mayday_
     
    Last edited: Apr 13, 2008
  2. jcsd
  3. malawi_glenn

    malawi_glenn 4,727
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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. Q1

    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. malawi_glenn

    malawi_glenn 4,727
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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. 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. malawi_glenn

    malawi_glenn 4,727
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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. 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. malawi_glenn

    malawi_glenn 4,727
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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. 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.

    Q2

    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. I can find the momentum as I have both E and c, so I can do E/c
     
  12. 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. malawi_glenn

    malawi_glenn 4,727
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    no

    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. malawi_glenn

    malawi_glenn 4,727
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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. Malawi, I really have no idea at all!!
     
  16. malawi_glenn

    malawi_glenn 4,727
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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. 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

    Equations:
    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. malawi_glenn

    malawi_glenn 4,727
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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. Is a time variation mentioned in the given problem?
     
  20. malawi_glenn

    malawi_glenn 4,727
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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. My comments addressed only "Question 1" (constant power dissipation).
     
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