Question

What is the frequency and energy of electromagnetic radiation with a wavelength of `"1.22 km"`?

Answer 1

`f=2.459xx10^5 "Hz"`

`"E"=1.63xx10^(-28)"J"`

Explanation:

There are two parts to this question. The first part determines what the frequency will be. The second part determines what the energy will be.

Part 1: Determining Frequency

The speed of light is directly proportional to wavelength and frequency. `color(red)(c=lambda*f)`, where `c` is the speed of light, `lambda` is the wavelength, `f` is the frequency.

The speed of light `c` is a constant. `c=3.00xx10^8"m/s"` to three significant figures.

Since `c` is in `"m/s"`, the wavelength needs to be converted from kilometers to meters.

`1.22cancel"km"xx(1000"m")/(1 cancel"km")="1220 m"`

Now we're reading to determine the frequency.

Known/Given
`c=3.00xx10^8"m/s"`
`lambda="1220 m"`

Unknown
`f`

Solution
Rearrange the equation to isolate `f`, substitute the known/given values into the equation.

`f=c/lambda`

`f=(3.00xx10^8cancel"m"/"s")/(1220cancel"m")=(2.459xx10^5)/"s"=2.459xx10^5 "Hz"`

Part 2: Determine the Energy

The energy in a photon is determined by multiplying the frequency by Planck's constant. Planck's constant, `h`, is `6.63xx10^(-34)"J"*"s"`. The unit for energy is the Joule `("J")`.

The equation for determining the energy of a photon is shown below.

`E=hf`, where `"E"` is energy, `h` is Planck's constant, and `f` is the frequency.

Substitute the known values into the equation and solve.

`E=(6.63xx10^(-34)"J"*cancel"s")((2.459xx10^5)/cancel"s")=1.63xx10^(-28)"J"`