Question

Part C Write a balanced nuclear equations for the formation of the isotope having 177 neutrons. Express your answer as a nuclear equation. 0 0 28 Ca +247 Bk 294 Uus tn Ca +247 Bk 24 Uus + 25n O 8Ca +387 Bk-4Uus +ổn 28 Ca +247 Bk 204 Uus + 3n

Answer 1

In a balanced nuclear equation both sideste reacts and products) number of total protons e number of mass number is equal. 48 ca + 247 BK 20 294 Uus tn 117 Reactants Products side side number of protons = 20+97= 117 number of protons = 117 to = 117 Reactans side mass number = 48+247 = 295 Products side mass number=294+1=295 .. This is balanced nuclear equation number of heutrons = mass numbex - protons 294-117 - 177 heutrong ® 48ca +247 Bk 294 uus ta 'n 20 qy Bk Reactants side number of protons : 20+97-117 products side iwe number number of protons = of protons = 1 17+2(0) = 117 peactants side mass number = 48 +247 = 295 products side mass number = 294+2(1) = 296 mass number not equal A not a balanced equation

* Hoteca + 2470K saat dus + in reactants side number of profons = 20 +98 = 118 Produch side number of protons = 117 +0 =117 Protons are not equal a not a balanced equation 48ca +247 BK - 20 98 294, TISUUS +315 117 uso" Reactants side number of protons = 20 +98 = 118 produch side number of protons = 117+ 3(0) = 117 protons are not equal is not a balanced equation.