Question

1. the emission spectrum of mercury atoms has a bright green line with wavelength 486.1 nm. a. calculate the frequency of these photons. show your work.

Explanation:

Step1: Recall the speed - of - light formula

The speed of light formula is $c = \lambda
u$, where $c$ is the speed of light ($c = 3\times10^{8}\ m/s$), $\lambda$ is the wavelength, and $
u$ is the frequency. We need to solve for $
u$, so $
u=\frac{c}{\lambda}$.

Step2: Convert the wavelength to SI units

The given wavelength $\lambda = 486.1\ nm$. Since $1\ nm=1\times 10^{-9}\ m$, then $\lambda = 486.1\times10^{-9}\ m$.

Step3: Calculate the frequency

Substitute $c = 3\times10^{8}\ m/s$ and $\lambda = 486.1\times 10^{-9}\ m$ into the formula $
u=\frac{c}{\lambda}$.
$
u=\frac{3\times 10^{8}\ m/s}{486.1\times 10^{-9}\ m}\approx6.17\times 10^{14}\ Hz$

Answer:

$6.17\times 10^{14}\ Hz$