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

[zeromodz]

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?

 

[Integral]

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

 

[Matterwave]

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.

 

[Ich]

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.

 

[atyy]

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)

 

[ideasrule]

To avoid confusion, I'd use "f".

 

[arunma]

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:

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