If light has a wavelength of 468 nm, what is the energy of this light expressed with units of kJ/mol?

Jun 4, 2017

$256$ $\frac{k J}{m o l}$

Explanation:

Think of light as travelling in discrete packets of energy called photons. The energy of a photon is given by the Planck-Einstein relation, which states that the energy of a photon is equal to its frequency ($\nu$) multiplied by a proportionality constant ($h$), known as Planck's constant:

${E}_{\text{photon}} = h \cdot \nu$

where

$h = 6.63 \cdot {10}^{-} 34 \text{ J"*"s}$

The frequency of a photon is related to its wavelength and is given by the speed of light, divided by the wavelength:

$\nu = \frac{c}{\lambda}$

Now substitute this into the energy expression:

${E}_{\text{photon}} = \frac{h \cdot c}{\lambda}$

The number of particles in one mole is given by the Avogadro constant, which is ${N}_{A} = 6.02 \cdot {10}^{23}$ $m o {l}^{-} 1$. In this case the particles are photons.
As we have the expression for energy per photon, we need to multiply it by the Avogadro constant to get the amount of energy per one mole of photons.

${E}_{\text{per mole}} = \frac{{N}_{A} \cdot h \cdot c}{\lambda}$

Note how the units for this expression cancel out to what we desire:

$\frac{J \cdot s \cdot m}{s \cdot m \cdot m o l} = \frac{J}{m o l}$

Using SI units ($m$, $\frac{m}{s}$, $J \cdot s$, $m o {l}^{-} 1$), substitute all the values into the equation to get the answer:

${E}_{\text{per mole}} = \frac{6.02 \cdot {10}^{23} \cdot 6.63 \cdot {10}^{-} 34 \cdot 3.00 \cdot {10}^{8}}{468 \cdot {10}^{-} 9}$
$= 256 \cdot {10}^{3}$ $\text{ J"/"mol}$

Finally, convert joules to kilojoules:

$E = 256$ $\text{ kJ"/"mol}$