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

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SUMMARY

The equation C = Vλ describes the relationship between the speed of light (c), frequency (ν), and wavelength (λ) of a photon. For a photon with a wavelength of 450nm, the frequency is calculated using ν = c/λ, resulting in approximately 7 x 1014 Hz, not the previously calculated 6.66 x 1018 Hz. To find the energy of the photon, the formula E = hν can be used, where h is Planck's constant (6.626 x 10-34 J·s). An alternative formula for energy in electron volts is E = 1240 eV·nm/λ when λ is expressed in nanometers.

PREREQUISITES
  • Understanding of the speed of light (c = 299,792,458 m/s)
  • Familiarity with Planck's constant (h = 6.626 x 10-34 J·s)
  • Knowledge of wavelength and frequency conversion (e.g., 450nm = 4.5 x 10-7 m)
  • Basic grasp of dimensional analysis in physics
NEXT STEPS
  • Study the derivation of the energy formula E = hν
  • Learn about the significance of Planck's constant in quantum mechanics
  • Explore the concept of photon energy in different units (e.g., Joules vs. electron volts)
  • Investigate the implications of wavelength on photon energy and frequency
USEFUL FOR

Students in physics, educators teaching quantum mechanics, and anyone interested in the properties of light and photons will benefit from this discussion.

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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