# What is the wavelength of a photon of blue light whose frequency is #6.3 * 10^14# #s^-1#?

##### 1 Answer

#### Explanation:

The key to any frequency and wavelength problem is the fact that frequency and wavelength have an **inverse relationship** described by the equation

#color(blue)( lamda * nu = c)" "# , where

So, what does an *inverse relationship* mean?

In simple words, that equation tells you that when frequency **increases**, wavelength **must decrease** in order for their product to remain constant.

Likewise, when frequency **decreases**, wavelength **must increase** in order for their product to remain constant.

As a consequence of this, you can expect waves that have *short wavelengths* to also have *high* frequencies, and waves that have *long* wavelengths to also have *short* frequencies.

In your case, the wave has a relatively *high frequency*, so you can predict a relatively short wavelength.

Another important thing to notice in such problems is the fact that the **units** used to express the frequency **match** those used to express the speed of light.

This tells you that you don't have to do any unit conversions.

So, plug in your values to get

#lamda * nu = c implies lamda = c/(nu)#

#lamda = (3 * 10^8 "m" color(red)(cancel(color(black)("s"^(-1)))))/(6.3 * 10^14 color(red)(cancel(color(black)("s"^(-1))))) = color(green)(4.8 * 10^(-7)"m"#

The answer is rounded to two sig figs. You can express this wavelength in *nanometers* by using the conversion factor

#"1 m" = 10^9"nm"#

This will give you

This wavelength places your wave in the *visible part* of the **electromagnetic spectrum**. More specifically, this wave would be located in the *blue region* of the spectrum.