Question 6.4.Q3: Several empirical expressions have been proposed for calcula......

Before using (6.25) we must determine the mean atomic ionization/excitation potential for the following elements: hydrogen and oxygen as constituents of water \left(\mathrm{H}_2 \mathrm{O}\right) and hydrogen, carbon, and oxygen as constituents of Lucite \left(\mathrm{C}_5 \mathrm{H}_8 \mathrm{O}_2\right)_n. The appropriate data are presented in Table 6.7.

(a) Mean ionization/excitation potential of water is calculated from (6.25) as follows: We first calculate the two components of (6.25) and get

\sum_i N_i Z_i \ln I_i=2 \times 1 \times \ln 19+1 \times 8 \times \ln 104.6=43.1           (6.26)

and

\sum_i N_i Z_i=2 \times 1+1 \times 8=10             (6.27)

and then use (6.25) to get

\ln I=\frac{\sum_i N_i Z_i \ln I_i}{\sum_i N_i Z_i}=\frac{43.1}{10}=4.31 \quad \text { or } \quad I=74.4 \mathrm{eV}            (6.28)

in good agreement with the value of 75 eV that is in common use for the mean ionization/excitation potential of water.

(b) Mean ionization/excitation potential I of Lucite \left(\mathrm{C}_5 \mathrm{H}_8 \mathrm{O}_2\right)_n is also calculated from (6.25) by first calculating the numerator and denominator of (6.25), respectively

\sum_i N_i Z_i \ln I_i=5 \times 6 \times \ln 81.5+8 \times 1 \times \ln 19+2 \times 8 \times \ln 104.6=230            (6.29)

and

\sum_i N_i Z_i=5 \times 6+8 \times 1+2 \times 8=54              (6.30)

and then using (6.25) to get

\ln I=\frac{\sum_i N_i Z_i \ln I_i}{\sum_i N_i Z_i}=\frac{230}{54}=4.26            (6.31)

resulting in

I=e^{4.26}=71 \mathrm{eV}             (6.32)

in reasonable agreement with the value of 74 eV that is commonly used for the ionization/excitation potential of Lucite.