using System; using L=Science.Physics.GeneralPhysics; namespace Serway.Chapter44 { /// /// Example08: The Energy Liberated when Radium Decays /// The 226Ra nucleus undergoes alpha decay according to Equation /// 44.11. Calculate the value for this process. Take the /// masses to be 226.025403 u for 226Ra, 222.017570 u for 222Rn, /// and 4.022603 u for 4He, as found in Table A.3. /// public class Example08 { public Example08() { } private string result; public string Result { get{return result;} } public void Compute() { L.Decay obj = new L.Decay(); L.Nucleus Ra = new L.Nucleus(); Ra.MassNumber = 226; Ra.AtomicNumber = 88; Ra.NeutronNumber = 226 - 88; L.Nucleus X = L.Decay.AlphaDecay(Ra); result+=Convert.ToString(X.AtomicNumber); } } } //86