Chapter 7 - Topic 7A - The origins of quantum mechanics - Exercises - Page 310: 7A.2(b)

$a) $ $E_{photon} = 9.939 \times 10^{-19} J$ $E_{mole} = 598547 \space Jmol^{-1} = 598.547 \space kJ mol^{-1}$ $b) $ $E_{photon} = 1.325 \times 10^{-15} J$ $E_{mole} = 798\times 10^6 \space Jmol^{-1} = 798 \times 10^3 \space kJ mol^{-1}$ $c) $ $E_{photon} = 1.99 \times 10^{-23} J$ $E_{mole} = 11.97 \space Jmol^{-1} =0.012\space kJ mol^{-1}$

Energy per photon is given by i.e $E_{photon} = {hc \over \lambda}$ Energy per mole is given by i.e $E_{mole} = N_A E_{photon}$ where $N_A$ is Avagadro's number equal to $N_A = 6.022 \times 10^{23} mol^{-1}$ $a) \space \lambda = 200 nm$ $E_{photon} = {hc \over \lambda} = {6.626 \times 10^{-34} \times 3 \times 10^8 \over 200 \times 10^{-9}} = 9.939 \times 10^{-19} J$ $E_{mole} = N_A E_photon = 6.022 \times 10^{23} \times 9.939\times 10^{-19} J mol^{-1} $ $E_{mole} = 598547 \space Jmol^{-1} = 598.547 \space kJ mol^{-1}$ $b) \space \lambda = 150 pm$ $E_{photon} = {hc \over \lambda} = {6.626 \times 10^{-34} \times 3 \times 10^8 \over 150 \times 10^{-12}} = 1.325 \times 10^{-15} J$ $E_{mole} = N_A E_photon = 6.022 \times 10^{23} \times 1.325\times 10^{-15} J mol^{-1} $ $E_{mole} = 798\times 10^6 \space Jmol^{-1} = 798\times 10^3 \space kJ mol^{-1}$ $c) \space \lambda = 1 cm$ $E_{photon} = {hc \over \lambda} = {6.626 \times 10^{-34} \times 3 \times 10^8 \over 1 \times 10^{-2}} = 1.99 \times 10^{-23} J$ $E_{mole} = N_A E_photon = 6.022 \times 10^{23} \times 1.99\times 10^{-23} J mol^{-1} $ $E_{mole} = 11.97 \space Jmol^{-1} =0.012\space kJ mol^{-1}$