# How do I convert between wavelength and frequency and wavenumber?

Aug 18, 2015

I assume you are referring to interconverting between $\lambda$, $\nu$, and $t i l \mathrm{de} \nu$, or something like that? I say this because $\lambda$ is typically wavelength in $\text{nm}$, $\nu$ is typically frequency in ${\text{s}}^{-} 1$, and $t i l \mathrm{de} \nu$ means energy in wavenumbers (${\text{cm}}^{-} 1$). If you want ${\text{nm}}^{- 1}$, just reciprocate $\text{nm}$.

There are four possibilities for conversions that I could cover:

1. $\lambda \to \nu$
2. $\nu \to t i l \mathrm{de} \nu$
3. $\nu \to \lambda$
4. $t i l \mathrm{de} \nu \to \nu$

However, recognize that if you can do 1 and 2, you have done 3 and 4 backwards, and if you can do 1 and 2 consecutively, you can go straight from $\lambda$ to $t i l \mathrm{de} \nu$ (same with 3 and 4 but $t i l \mathrm{de} \nu$ to $\lambda$). So, I will only show 1 and 2.

• $\lambda \to \nu$

Suppose we have $\lambda = 600 n m$ for yellow light and we want its frequency in ${s}^{-} 1$.

What we want is to convert from a unit of length to a unit of reciprocal time, which requires something that has $\text{length"/"time}$ units... The speed of light works great here, and it's about $3 \times {10}^{8} \text{m/s}$. Therefore:

1. Reciprocate the wavelength
2. Convert to $\text{m}$
3. Multiply by the speed of light

overbrace((1/(600 cancel("nm"))))^(lambda)xx((10^9 cancel("nm"))/(1 cancel("m")))xx(3xx10^8 cancel("m")/"s") = underbrace(color(blue)(5.bar55xx10^(-3) "s"^-1))_(nu)

• $\nu \to t i l \mathrm{de} \nu$

This is fairly straightforward. We have $\frac{1}{\text{s}}$ and want $\frac{1}{\text{cm}}$. Suppose we have a frequency of $6 \times {10}^{- 3} {\text{s}}^{-} 1$.

1. Divide by the speed of light
2. Convert to $\text{cm}$

overbrace((6xx10^(-3) 1/cancel("s")))^(nu)xx(cancel("s")/(3xx10^8 cancel("m")))xx((1 cancel("m"))/(100 "cm")) = underbrace(color(blue)(2xx10^(-13) "cm"^(-1)))_(tildenu)