Question

what is the energy of a mole of photons that have a wavelength of 615 nm?

Explanation:

Step1: Recall the energy - wavelength formula for a single photon

The energy of a single photon is given by $E = h
u$, and since $
u=\frac{c}{\lambda}$, then $E=\frac{hc}{\lambda}$, where $h = 6.626\times10^{-34}\ J\cdot s$ (Planck's constant), $c = 3.0\times10^{8}\ m/s$ (speed of light) and $\lambda$ is the wavelength. First, convert the wavelength $\lambda=615\ nm = 615\times10^{-9}\ m$.
$E_{photon}=\frac{6.626\times10^{-34}\ J\cdot s\times3.0\times10^{8}\ m/s}{615\times10^{-9}\ m}$

Step2: Calculate the energy of a single photon

$E_{photon}=\frac{6.626\times10^{-34}\times3.0\times10^{8}}{615\times10^{-9}}\ J\approx3.23\times 10^{-19}\ J$

Step3: Calculate the energy of one - mole of photons

One mole of photons contains $N = N_{A}=6.022\times10^{23}$ photons (Avogadro's number). The energy of one - mole of photons $E_{mole}=N_{A}\times E_{photon}$.
$E_{mole}=6.022\times10^{23}\times3.23\times10^{-19}\ J$
$E_{mole}=6.022\times3.23\times10^{23 - 19}\ J\approx194000\ J=194\ kJ$

Answer:

$194\ kJ$