Spectral Intensity Formula by Using Frequency

In summary, the conversation discusses using Pointon's book to find the intensity by using frequency. The Doppler effect is mentioned, and the formula for frequency is revised to use 'F' instead of ##\omega##. The next step is to determine the Doppler shifted frequency based on the velocity of the source. The conversation also notes that the mass of the element will need to be left as a parameter in the equation and suggests using 'g' for the distribution function to avoid confusion with frequency denoted by 'f'.
  • #1
benagastov
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0
Homework Statement
It is known that a hypothetical atom emits light when it transitions from an excited state to a ground state with a frequency w0 = 8 x 10^14 Hz. It is known that the observer receives the frequency of these photons due to the Doppler effect.

Get the intensity of the photon spectrum!
Relevant Equations
\omega \:=\:\omega _0\cdot \left(1\:\pm \frac{v_x}{c}\right)

w = w0 (1 + vx/c)
I found there is kind of solution in Pointon's book: An Introduction to Statistical Physics for Students. But I don't know how to find intensity by using frequency.
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  • #2
Since no one has replied yet, see if this helps.

benagastov said:
It is known that the observer receives the frequency of these photons due to the Doppler effect.
That's not clear. I guess you mean:
“Since the atoms are moving ramdomly, due to the Doppler effect the observer receives photons with a distribution of frequencies”.

benagastov said:
w = w0 (1 + vx/c)
##\omega## is generally used for angular frequency. 'f' is already used for the distribution function. ##\nu## (Greek letter 'nu') looks too much like 'v' (velocity symbol).

Maybe use 'F' for frequency.

benagastov said:
I found there is kind of solution in Pointon's book: An Introduction to Statistical Physics for Students. But I don't know how to find intensity by using frequency.

View attachment 299395
Since ##\lambda= \frac c F## some simple algebra allows you to rewrite equation 3.23a (from your attachment) in terms of frequency.

Then you can use the same method as shown in your attachment.

Edited to provide only initial guidance.
 
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  • #3
Steve4Physics said:
Since no one has replied yet, see if this helps.That's not clear. I guess you mean:
“Since the atoms are moving ramdomly, due to the Doppler effect the observer receives photons with a distribution of frequencies”.##\omega## is generally used for angular frequency. 'f' is already used for the distribution function. ##\nu## (Greek letter 'nu') looks too much like 'v' (velocity symbol).

Maybe use 'F' for frequency.Since ##\lambda= \frac c F## some simple algebra allows you to rewrite equation 3.23a (from your attachment) in terms of frequency.

Then you can use the same method as shown in your attachment.

Edited to provide only initial guidance.
Thank you for your reply! Is this how I do it? I use Bose_einstein statistics since it was Photon
 

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  • #4
benagastov said:
Thank you for your reply! Is this how I do it? I use Bose_einstein statistics since it was Photon
I think you have misunderstood the physics involved.

This is primarily a classical system. The atoms, if stationary, would emit waves of frequency 8x10¹⁴ Hz.

But each atom is a moving source with a velocity component in the observed x-direction of ##v_x##. The Doppler shift is due to having having moving sources (moving atoms). That’s how a distribution of frequencies arises.

The atoms are assumed to have a (classical) Maxwell-Boltzmann velocity distribution. So we know the distribution of the ##v_x##s.

Bose-Einstein statistics are not relevant in the context of this question.

If you don’t understand this, you need to:
- revise the Doppler effect;
- revise the Maxwell-Boltzmann velocity distribution;
- make sure you fully understand the working used in your Post #1 attachment.

You have ignored the instructions I gave in Post #2. So I will repeat them for you:
"Since ##\lambda = \frac c F## some simple algebra allows you to rewrite equation 3.23a (from your attachment) in terms of frequency.

Then you can use the same method as shown in your attachment.”

So your first task is to produce a new version of equation 3.23a with frequency instead of wavelength.

(Also, consider using ‘F’ rather than ##\omega## for frequency, for the reasons I’ve already explained.)
 
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  • #5
Ok, I am sorry took so long. After I discharged from hospital, I took rest time.
So I think I already finish the formula, is this Blue Shift? What I do next? There is no mass in my question.

Problem.PNG
 
  • #6
benagastov said:
So I think I already finish the formula, is this Blue Shift?
No. The formula (you highlighted in green) gives the Doppler shifted frequency.

If ##v_x## is negative:
- is the source (the atom) moving towards or away from the observer?
- is the observed frequency bigger or smaller than ##f_0##?
- is this a blue shift or a red shift?

If ##v_x## is positive:
- is the source (the atom) moving towards or away from the observer?
- is the observed frequency bigger or smaller than ##f_0##?
- is this a blue shift or a red shift?

You have (correctly) found ##v_x = \frac {c(f-f_0)}{f}##. But you then say ##dv_x = \frac {cf_0}{f} df##. That’s wrong. Can you correct it?

When you have have the correct expression for ##dv_x## you can use it in the distribution function.

benagastov said:
What I do next? There is no mass in my question.
You will have to leave 'm' (mass of an atom of the unspecified element) as a parameter in the equation.

As previously noted, using ##\omega## is confusing as it is usually used for angular frequency.

Maybe you could use ‘g’ for the distribution function rather than ‘f’? Then you could use f for frequency.
 
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1. What is the spectral intensity formula by using frequency?

The spectral intensity formula by using frequency is a mathematical equation that describes the relationship between the intensity of light and its corresponding frequency. It is commonly used in the field of optics and spectroscopy to analyze the properties of light.

2. How is the spectral intensity formula derived?

The spectral intensity formula is derived from the Planck's law of blackbody radiation, which states that the intensity of radiation emitted by a blackbody is directly proportional to the frequency of the radiation and the temperature of the body.

3. What are the units of measurement for spectral intensity?

The units of measurement for spectral intensity are watts per square meter per Hertz (W/m²/Hz) or joules per second per square meter per Hertz (J/s/m²/Hz).

4. How is the spectral intensity formula used in practical applications?

The spectral intensity formula is used in various practical applications, such as in the design of optical instruments, analysis of light sources, and in spectroscopy to determine the composition of materials. It is also used in the study of astronomical objects and in the development of new technologies, such as lasers and LEDs.

5. Are there any limitations to the spectral intensity formula?

One limitation of the spectral intensity formula is that it assumes the light source to be a perfect blackbody, which is not always the case in real-world scenarios. It also does not take into account other factors such as scattering and absorption of light, which can affect the overall intensity. Additionally, the formula may not be applicable to all types of electromagnetic radiation, as it is specifically derived for thermal radiation.

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