Question: Calculate Wavelength from Intensity

AI Thread Summary
To calculate the wavelength of the most intensive microwaves, the intensity is given as 380 units with 5 waves per centimeter. The discussion highlights the need to understand the relationship between intensity and photon energy, referencing the formula Wavelength = HC/E. Participants express confusion about determining the energy (E) of a photon to apply this formula effectively. Clarification is sought regarding the meaning of "units" from the graph used to derive the intensity. Understanding these concepts is crucial for solving the wavelength calculation problem.
Hiyun
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Homework Statement


Question: Calculate the wavelength in mm of the most intensive microwaves

Question before: State the intensity of the most intensive microwaves detected
Answer: 380 Units. (5 Waves per CM)
[Read from a graph]

Homework Equations



None given.

The Attempt at a Solution



I have never attempted a question to calculate wavelength from intensity. After some quick research on the internet, information about photons etc which I had never come across before showed up. If anyone would like to explain, i'd be very grateful.
 
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Hiyun said:

Homework Statement


Question: Calculate the wavelength in mm of the most intensive microwaves

Question before: State the intensity of the most intensive microwaves detected
Answer: 380 Units. (5 Waves per CM)
[Read from a graph]



Homework Equations



None given.

The Attempt at a Solution



I have never attempted a question to calculate wavelength from intensity. After some quick research on the internet, information about photons etc which I had never come across before showed up. If anyone would like to explain, i'd be very grateful.

This sounds like a microwave equivalent of a common optics question.

There is a formula of calculating the energy associated with a photon of Red light, and a photon of blue light (one is bigger than the other).

Thus if you have a 6W red light and a 6W blue light (6 joules given off each second), then there will be different numbers of photons given off by each light.
If one phonon has 0.5 J of energy, you need 12 per second, while if the other photon has 0.6 J of energy, you need only 10 per second.

Perhaps that example/concept will help.
 
I did more research into this and I found out that Wavelength=HC/E

But I'm still stuck on getting the E (energy of a photon)

So far I've got

WL=1.99x10-25 J-m x E
 
Hiyun said:
I did more research into this and I found out that Wavelength=HC/E

But I'm still stuck on getting the E (energy of a photon)

So far I've got

WL=1.99x10-25 J-m x E

In your original post you said the nswer to the previous question was
Answer: 380 Units. (5 Waves per CM) which you got from a graph.

Using the scale of that graph what is a "unit"?
 
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