# Anyone care to explain the TA's note? (1 Viewer)

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#### Shackleford

Omega here is frequency.

I used lambda = c/f. Why is it (2*pi*c)/omega and not c/f or c/omega?

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-10-04203213.jpg?t=1286245112 [Broken]

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-10-04203118.jpg?t=1286245065 [Broken]

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#### fzero

Homework Helper
Gold Member
It looks like the formula you used was derived as

$$\omega_{ij} = \frac{E_j - E_i}{\hbar},$$

which is an angular frequency as can be seen from the presence of $$\hbar$$ rather than $$h$$.

#### Shackleford

It looks like the formula you used was derived as

$$\omega_{ij} = \frac{E_j - E_i}{\hbar},$$

which is an angular frequency as can be seen from the presence of $$\hbar$$ rather than $$h$$.
Yeah. But h-bar is already in the denominator in the formula I used to calculate $$omega_{ij}$$.

#### fzero

Homework Helper
Gold Member
The wavelength is still $$2\pi c/\omega$$.

#### cristo

Staff Emeritus
Why not ask your TA for clarification?

#### Shackleford

The wavelength is still $$2\pi c/\omega$$.
Why is lambda = c/f not valid?

#### Shackleford

Why not ask your TA for clarification?
I forgot. I already bugged him about two other problems. I did poorly on this damn homework.

#### fzero

Homework Helper
Gold Member
Why is lambda = c/f not valid?
It is, but $$\omega = 2\pi f$$.

#### Shackleford

It is, but $$\omega = 2\pi f$$.
Well, that's my mistake. I wasn't taking omega to be an angular frequency. Why? I don't know. Son of a *****.

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