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

$a) $ $E_{photon} = 3.313 \times 10^{-19} J$ $E_{mole} = 199513 \space Jmol^{-1} = 199.513 \space kJ mol^{-1}$ $b) $ $E_{photon} = 3.614 \times 10^{-19} J$ $E_{mole} = 217653 \space Jmol^{-1} = 217.653 \space kJ mol^{-1}$ $c) $ $E_{photon} = 4.97 \times 10^{-19} J$ $E_{mole} = 299273 \space Jmol^{-1} =299.273\space kJ mol^{-1}$

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 $N_A = 6.022 \times 10^{23} mol^{-1}$ $a) \space \lambda = 600 nm$ $E_{photon} = {hc \over \lambda} = {6.626 \times 10^{-34} \times 3 \times 10^8 \over 600 \times 10^{-9}} = 3.313 \times 10^{-19} J$ $E_{mole} = N_A E_photon = 6.022 \times 10^{23} \times 3.313\times 10^{-19} J mol^{-1} $ $E_{mole} = 199513 \space Jmol^{-1} = 199.513 \space kJ mol^{-1}$ $b) \space \lambda = 550 nm$ $E_{photon} = {hc \over \lambda} = {6.626 \times 10^{-34} \times 3 \times 10^8 \over 550 \times 10^{-9}} = 3.614 \times 10^{-19} J$ $E_{mole} = N_A E_photon = 6.022 \times 10^{23} \times 3.614\times 10^{-19} J mol^{-1} $ $E_{mole} = 217653 \space Jmol^{-1} = 217.653 \space kJ mol^{-1}$ $c) \space \lambda = 400 nm$ $E_{photon} = {hc \over \lambda} = {6.626 \times 10^{-34} \times 3 \times 10^8 \over 400 \times 10^{-9}} = 4.97 \times 10^{-19} J$ $E_{mole} = N_A E_photon = 6.022 \times 10^{23} \times 4.97\times 10^{-19} J mol^{-1} $ $E_{mole} = 299273 \space Jmol^{-1} =299.273\space kJ mol^{-1}$