Question

the green light emitted by a stoplight has a wavelength of 519 nm. what is the frequency of this photon? (c

Answer 1

Approximately
`1.73 * 10^(-15)\; {\rm s^(-1)}` (or equivalently,
`1.73 * 10^(-15)\; {\rm Hz}`) assuming that the wavelength was measured in vacuum.

Step-by-step explanation:

Look up the speed of light in vacuum:
`v \approx 3.00 * 10^(8)\; {\rm m\cdot s^(-1)}`.

Wavelength
`\lambda` is the distance the wave travelled in one period
`T` of time. Hence, multiplying period by the speed
`v` of the wave will give the wavelength of the wave:
`\lambda = v\, T`. Rearrange to obtain
`T = (\lambda / v)`.

The frequency
`f` of a wave is the number of periods of this wave in unit time. Frequency is equal to the reciprocal of period:
`f = (1/T)`. However, since
`T = (\lambda / v)`:

`\begin{aligned}f &= (1)/(T) \\ &= (1)/(\lambda / v) \\ &= (v)/(\lambda)\end{aligned}`.

Apply unit conversion and ensure that the wavelength of this wave is measured in standard units:

`\begin{aligned}\lambda &= (519\; {\rm nm})\, \frac{10^(-9)\; {\rm m}}{1\; {\rm nm}} \\ &= 5.19 * 10^(-7)\; {\rm m}\end{aligned}`.

The speed of the wave in this question is equal to the speed of light in vacuum,
`v \approx 3.00 * 10^(8)\; {\rm m\cdot s^(-1)}`. Substitute in the value of wave speed
`v` and wavelength
`\lambda` to find the frequency
`f` of this wave:

`\begin{aligned}f &= (v)/(\lambda) \\ &\approx \frac{3.00* 10^(8)\; {\rm m\cdot s^(-1)}}{5.19 * 10^(-7)\; {\rm m}} \\ &\approx 1.73 * 10^(-15)\; {\rm s^(-1)} \\ &= 1.73* 10^(-15)\; {\rm Hz}\end{aligned}`.