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diff --git a/3845/CH32/EX32.1/Ex32_1.sce b/3845/CH32/EX32.1/Ex32_1.sce
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+//Example 32.1

+//Activity R=1.00*10^-6 Ci=3.70*10^4 Bq=3.70*10^4 decays/s

+m=2;//Mass (kg)

+RBE=20;//Relative Biological Effectiveness, See Table 32.2

+d=(3.70*10^4)*(3.16*10^7);//Number of decays per year=R*Number of seconds in a year

+E=d*5.23*1.60*10^-13;//Ionisation energy deposited per year (J)

+Dose1=(E/m)/1;//Dose (Gy)

+Dose2=Dose1*20;//Dose (Sv)

+printf('Dose in Gy = %0.3f Gy',Dose1)

+printf('\nDose in Sv = %0.1f Sv',Dose2)

+//Openstax - College Physics

+//Download for free at http://cnx.org/content/col11406/latest

+

diff --git a/3845/CH32/EX32.2/Ex32_2.sce b/3845/CH32/EX32.2/Ex32_2.sce
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+//Example 32.2

+a1=2.014102;//Atomic mass of deuterium (u), See Appendix A

+a2=3.016050;//Atomic mass of tritium (u), See Appendix A

+a=a1+a2;//Total atomic mass per reaction (u)

+m=a;//Mass per mole of reactants (g/mol)

+mol_of_reactants=1000/m;//Mol of reactants in 1kg

+N=mol_of_reactants*6.02*10^23;//Number of reactions

+E=N*17.59*(1.602*10^-13);//Total energy output=number of reactions*energy per reaction) (J)

+printf('a. Total energy output = %0.2e J',E)

+t=3.16*10^7;//Number of seconds in a year (s)

+power=E/t;

+printf('\nb. Power output = %0.1f MW',power/10^6)

+//There is a small variation in the value of atomic mass of Tritium used in the textbook from that found in Appendix A

+//Openstax - College Physics

+//Download for free at http://cnx.org/content/col11406/latest

+

diff --git a/3845/CH32/EX32.3/Ex32_3.sce b/3845/CH32/EX32.3/Ex32_3.sce
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+//Example 32.3

+m_prod=94.919388+139.921610+3*(1.008665);//Total mass of products (Sr 95, Xe 140, n)(u)

+delta_m=238.050784-m_prod;//Mass lost (u)

+E=delta_m*931.5;//Energy released (MeV)

+//E=delta_m*(931.5MeV/c^2)/u*c^2

+printf('Energy released = %0.1f MeV',E)

+//Openstax - College Physics

+//Download for free at http://cnx.org/content/col11406/latest

+

diff --git a/3845/CH32/EX32.4/Ex32_4.sce b/3845/CH32/EX32.4/Ex32_4.sce
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+//Example 32.4

+m=235.04;//Mass of one mole of 235 U (g)

+mol=1000/m;//Number of moles in 1 kg (mol)

+atoms=mol*6.02*10^23;//Number of atoms in 1kg of 235 U

+E=atoms*200*1.6*10^-13;//Total energy released (J)

+printf('Total energy released = %0.2e J',E)

+//The answer varies due to round off error

+//Openstax - College Physics

+//Download for free at http://cnx.org/content/col11406/latest