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

Explanation:

Step1: Recall energy - wavelength formula for a single photon

The energy of a single photon is given by $E = h\nu$, and since $\nu=\frac{c}{\lambda}$, 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 from $nm$ to $m$: $\lambda=615\ nm = 615\times10^{-9}\ m$.

Step2: Calculate the energy of a single photon

$E=\frac{hc}{\lambda}=\frac{(6.626\times 10^{-34}\ J\cdot s)\times(3.0\times 10^{8}\ m/s)}{615\times 10^{-9}\ m}$ $E=\frac{19.878\times10^{-26}\ J\cdot m}{615\times 10^{-9}\ m}\approx 3.23\times 10^{-19}\ J$

Step3: Calculate the energy of a mole of photons

We know that 1 mole contains $N_A = 6.022\times10^{23}$ photons. The energy of a mole of photons $E_{mol}$ is $E_{mol}=N_A\times E$. $E_{mol}=(6.022\times 10^{23})\times(3.23\times 10^{-19}\ J)$ $E_{mol}=6.022\times3.23\times10^{23 - 19}\ J$ $E_{mol}\approx194\times10^{3}\ J = 194\ kJ$

Answer:

$194\ kJ$