Verified Answer :
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Use the following relation: dE = h ( f2 - f1), where dE is the difference in the energy, h is the Planck's Constant and f1, f2 are the given frequencies.
dE = 6.626 X 10^ -34 ( 589.6 - 589) = 3.9756e-34
Answer : The energy difference for one mole of photon is 206.55 J/mol.
Explanation :
Formula used :
\(E=\frac{hc}{\lambda}\)
For two different wavelength the formula will be:
\(E_1=\frac{hc}{\lambda_1}\)
\(E_2=\frac{hc}{\lambda_2}\)
\(\Delta E=E_1-E_2=\frac{hc}{\lambda_1}-\frac{hc}{\lambda_2}\)
\(\Delta E=hc\times [\frac{1}{\lambda_1}-\frac{1}{\lambda_2}]\)
where,
\(\Delta E\) = difference in energy of photon = ?
h = Planck's constant = \(6.626\times 10^{-34}Js\)
c = speed of light = \(3\times 10^8m/s\)
\(\lambda_1\) = wavelength of photon 1 = 589.0 nm = \(589\times 10^{-9}m\)
conversion used : \(1nm=10^{-9}m\)
\(\lambda_2\) = wavelength of photon 2 = 589.6 nm = \(589.6\times 10^{-9}m\)
Now put all the given values in the above formula, we get:
\(\Delta E=(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)\times [\frac{1}{589\times 10^{-9}m}-\frac{1}{589.6\times 10^{-9}m}]\)
\(\Delta E=3.43\times 10^{-22}J\)
Energy difference for one mole of photon = \(\Delta E\times \text{Avogadro's number}\)
Energy difference for one mole of photon = \((3.43\times 10^{-22}J)\times (6.022\times 10^{23})=206.55J/mol\)
Therefore, the energy difference for one mole of photon is 206.55 J/mol.