Question

the green light emitted by a stoplight has a wavelength of 561 nm. what is the frequency, in s⁻¹, of this photon? (c = 3.00×10⁸ m/s). answer: hz

Explanation:

Step1: Convert wavelength to meters

Given $\lambda = 561\ nm=561\times10^{- 9}\ m$.

Step2: Use the formula $c = \lambda

u$
We know that the speed - of - light formula is $c=\lambda
u$, where $c = 3.00\times10^{8}\ m/s$ is the speed of light, $\lambda$ is the wavelength and $
u$ is the frequency. Rearranging for $
u$, we get $
u=\frac{c}{\lambda}$.
Substitute $c = 3.00\times10^{8}\ m/s$ and $\lambda = 561\times10^{-9}\ m$ into the formula: $
u=\frac{3.00\times10^{8}}{561\times10^{-9}}\ Hz$.

Step3: Calculate the frequency

$
u=\frac{3.00\times10^{8}}{561\times10^{-9}}=\frac{3.00\times10^{8 + 9}}{561}=\frac{3.00\times10^{17}}{561}\approx5.35\times10^{14}\ Hz$.

Answer:

$5.35\times10^{14}$