# Question about c = vλ

1. Dec 5, 2009

### 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?

Last edited: Dec 5, 2009
2. Dec 5, 2009

### Integral

Staff Emeritus
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

3. Dec 5, 2009

### 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.

4. Dec 7, 2009

### 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.

5. Dec 7, 2009

### atyy

Even more useless info:
http://en.wikipedia.org/wiki/Nu_(letter)

6. Dec 7, 2009

### ideasrule

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

7. Dec 7, 2009

### arunma

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:

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