How to Approach Questions Related to Waves

  • Thread starter _Mayday_
  • Start date
  • Tags
    Waves
In summary: E}{c}p = \frac{1.65\times10^{-18}}{3\times10^8} = 5.52\times10^{-27}I do not know what to do from here.
  • #1
_Mayday_
808
0
Hey, I'm having a few problems with the questions below, now I think a lot of it is to do with not knowing how to approach the question. Please forgive me if there is not a lot 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:
Physics news on Phys.org
  • #2
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:
  • #3
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.
 
  • #4
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]
?
 
  • #5
malawi_glenn said:
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]
?

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.
 
  • #6
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?
 
  • #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.
 
  • #8
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).
 
  • #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.

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:
  • #10
I can find the momentum as I have both E and c, so I can do E/c
 
  • #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...
 
  • #12
no

force = time derivative of momentum

You can't GATHER force, force is instanteous.

So if the bulb casts away 20W photons in the same direction (we was to assue it was a parallel 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
 
  • #13
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
 
  • #14
Malawi, I really have no idea at all!
 
  • #15
_Mayday_ said:
Malawi, I really have no idea at all!



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:
  • #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

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)
 
  • #17
no energy is not = Power times t!

P = dE/dt, it is different from the AVERAGE Power: P = E/t
 
  • #18
Is a time variation mentioned in the given problem?
 
  • #19
mikelepore said:
Is a time variation mentioned in the given problem?

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_
 
  • #20
My comments addressed only "Question 1" (constant power dissipation).
 
  • #21
ok great
 

1. What are the different types of waves?

There are several types of waves, including mechanical waves (such as sound waves and water waves) and electromagnetic waves (such as light and radio waves).

2. How are waves created?

Waves are created when a source of energy causes a disturbance in a medium, such as air, water, or the ground. This disturbance then travels through the medium, carrying the energy with it.

3. What is the relationship between wavelength, frequency, and amplitude?

Wavelength is the distance between two consecutive peaks or troughs of a wave, and is inversely proportional to frequency. Frequency is the number of waves that pass a certain point per second, and is directly proportional to energy. Amplitude is the maximum displacement of a wave from its resting position, and is directly proportional to energy.

4. How do waves interact with each other?

When two or more waves meet, they can either reinforce each other (constructive interference) or cancel each other out (destructive interference). The resulting wave depends on the amplitudes and wavelengths of the individual waves.

5. What is the Doppler effect and how does it apply to waves?

The Doppler effect is the apparent change in frequency of a wave when the source of the wave is moving relative to the observer. This can be observed with sound waves, where the pitch of a siren seems to change as an ambulance passes by, or with light waves, where the color of a star appears to shift as it moves closer or farther away from Earth.

Similar threads

  • Introductory Physics Homework Help
Replies
15
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
964
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
20
Views
2K
Replies
27
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
23
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
941
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
Back
Top