Question

1. the emission spectrum of mercury atoms has a bright green line with wavelength 486.1 nm. 1nm = 1×10⁻⁹m a. calculate the frequency of these photons. show your work. b. calculate the energy of one photon of this light (in joules). show your work. c. calculate the energy of one mole of these photos (in joules). show your work.

Explanation:

Step1: Recall the speed - wavelength - frequency relation

The speed of light $c = 3\times10^{8}\ m/s$, and the relation between speed of light $c$, wavelength $\lambda$ and frequency $
u$ is $c=\lambda
u$, so $
u=\frac{c}{\lambda}$. Given $\lambda = 486.1\ nm=486.1\times 10^{-9}\ m$.
$
u=\frac{3\times 10^{8}}{486.1\times 10^{-9}}=\frac{3\times 10^{8}}{4.861\times 10^{-7}}\approx6.17\times 10^{14}\ Hz$

Step2: Calculate the energy of one photon

The energy of a photon is given by $E = h
u$, and since $
u=\frac{c}{\lambda}$, we can also write $E=\frac{hc}{\lambda}$, where $h = 6.63\times 10^{-34}\ J\cdot s$.
$E=\frac{(6.63\times 10^{-34})\times(3\times 10^{8})}{486.1\times 10^{-9}}=\frac{19.89\times 10^{-26}}{4.861\times 10^{-7}}\approx4.09\times 10^{-19}\ J$

Step3: Calculate the energy of one - mole of photons

One mole of photons contains $N_A = 6.022\times 10^{23}$ photons. The energy of one - mole of photons $E_{mol}$ is $E_{mol}=N_A\times E$.
$E_{mol}=(6.022\times 10^{23})\times(4.09\times 10^{-19})=6.022\times4.09\times 10^{4}\approx2.46\times 10^{5}\ J$

Answer:

a. $6.17\times 10^{14}\ Hz$
b. $4.09\times 10^{-19}\ J$
c. $2.46\times 10^{5}\ J$