Question

consider a photon with wavelength 727 nm. what is the energy of a mole of photons of this wavelength (in kj)?

Explanation:

Step1: Convert wavelength to meters

Given wavelength \(\lambda = 727\ nm=727\times10^{-9}\ m\)

Step2: Use the formula \(E = hc/\lambda\) for energy of one photon

where \(h = 6.626\times 10^{-34}\ J\cdot s\) (Planck's constant) and \(c = 3\times 10^{8}\ m/s\) (speed of light)
\(E=\frac{6.626\times 10^{-34}\times3\times 10^{8}}{727\times 10^{-9}}\)
\(E=\frac{19.878\times 10^{-26}}{727\times 10^{-9}}\)
\(E = 2.734\times10^{-19}\ J\)

Step3: Calculate energy for one - mole of photons

Using \(N_{A}=6.022\times 10^{23}\ mol^{-1}\) (Avogadro's number)
\(E_{mole}=E\times N_{A}\)
\(E_{mole}=2.734\times 10^{-19}\times6.022\times 10^{23}\)
\(E_{mole}=1.647\times 10^{5}\ J\)

Step4: Convert energy to kilojoules

Since \(1\ kJ = 1000\ J\)
\(E_{mole}=\frac{1.647\times 10^{5}}{1000}\ kJ\)
\(E_{mole}=164.7\ kJ\)

Answer:

\(164.7\)