Answer "What Does C=Vλ Mean? Find Energy of Photon

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Discussion Overview

The discussion revolves around the equation C = Vλ, focusing on understanding its components and implications in the context of photon energy calculations. Participants explore the meanings of the variables involved, particularly in relation to wavelength and frequency, and how to derive the energy of a photon from these concepts.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant interprets C as the speed of light, V as the speed of frequency, and λ as wavelength, attempting to calculate frequency from a given wavelength of 450nm.
  • Another participant corrects the interpretation of the variables, stating that V should be represented as ν (nu), which denotes frequency, not speed.
  • Concerns are raised about the numerical calculation of frequency, with a participant indicating that the wavelength should be expressed as 450nm = 4.5 * 10^-7m, not 4.5 * 10^-11m.
  • There is a suggestion to use "f" instead of "v" to avoid confusion regarding the representation of frequency.
  • A later reply introduces the relationship between energy and frequency, suggesting that energy can be calculated using E = hν or E = hc/λ, and provides a formula for converting wavelength in nanometers to energy in electron volts.

Areas of Agreement / Disagreement

Participants generally agree on the need for clarity regarding the definitions of variables and the correct calculations. However, there is no consensus on the initial interpretation of the equation or the specific numerical values presented, indicating ongoing disagreement and refinement of ideas.

Contextual Notes

Participants express uncertainty about the correct interpretation of units and the implications of the exponential notation in calculations. There are also unresolved questions about the proper terminology for frequency and its representation.

zeromodz
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Okay, I just need to grasp what the equation C = Vλ means
.
Correct me if I am wrong.
c = speed of light
v = speed of frequency
λ = wavelength

So if I wanted to find the speed of the frequency of a photon with a wavelength of 450nm

So I would use
v = c/λ
v = 299,792,458/4.5 * 10^(-11)
= 6.662054622 * 10^18

Then, if I wanted to find the energy of the photon, I would just multiply that number by planks constant (6.26 * 10^-34) and I get the energy of a quantum leap from an atom and the photon's energy itself, right?
 
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You seem to have a couple of problems. First to find the units on a quantity do dimensional analysis. This means to do algebra on the units of the quantities involved. In this case we have

\nu = \frac c {\lambda}

the units of c are m/s the units of \lambda are m so you have

\nu = \frac {\frac m s} m

so the units for \nu are s -1 more commonly called Hertz or Hz.

Now for your numerical value, do you understand the meaning of the exponential term? Your number is off by many orders of magnitude. I get something like 7 e14 or 700,000,000,000,000Hz
 
450nm is 4.5*10^-7 not -11

Nano is 10^-9

Also, we usually just call Frequency "frequency" instead of "speed of frequency" since there's no real speed (in the physics sense) involved.
 
Integral already mentioned it, but explicitly: it's not v, it's \nu. While v normally denotes a speed, \nu stands for a frequency, which is quite different.
 
Ich said:
Integral already mentioned it, but explicitly: it's not v, it's \nu. While v normally denotes a speed, \nu stands for a frequency, which is quite different.

Even more useless info:
http://en.wikipedia.org/wiki/Nu_(letter)
 
To avoid confusion, I'd use "f".
 
zeromodz said:
Okay, I just need to grasp what the equation C = Vλ means
.
Correct me if I am wrong.
c = speed of light
v = speed of frequency
λ = wavelength

So if I wanted to find the speed of the frequency of a photon with a wavelength of 450nm

So I would use
v = c/λ
v = 299,792,458/4.5 * 10^(-11)
= 6.662054622 * 10^18

Then, if I wanted to find the energy of the photon, I would just multiply that number by planks constant (6.26 * 10^-34) and I get the energy of a quantum leap from an atom and the photon's energy itself, right?


In additon to what others mentioned...

Since you know that E=h\nu, you could just do this to find the energy of a photon:

\nu = \dfrac{c}{\lambda}

E = \dfrac{hc}{\lambda}

A helpful number to remember is the value of hc. So if you express your wavelength in nanometers, you can get the energy in electron volts like this:

E = \dfrac{1240eV\times nm}{\lambda}
 

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