Answer
a) $^4_2$He + $^{40}_{18}$Ar --> $^{43}_{19}$K + $^1_1$H
b) $^1_0$n + $^{238}_{92}$U --> $^{239}_{92}$U
c) $^1_0$n + $^{14}_7$N --> $^{14}_6$C + $^1_1$H
d) $^{209}_{83}$Bi + $^{64}_{28}$Ni --> $^{272}_{111}$Rg + $^1_0$n
Work Step by Step
a) To produce a potassium-43 isotope ( $^{43}_{19}$K) and a proton ( $^1_1$H), Argon-40 ($^{40}_{18}$Ar) must be bombarded with an alpha particle ($^4_2$He) to balance the equation.
b) To find the product of the bombardment of uranium-238 with a neutron, you would add 1 to 238, which would result in uranium-239.
c) To find the missing isotope, you would add the mass numbers of the two products, and the atomic numbers of the two products, respectively; then you would subtract 1 from the sum of the mass numbers and leave the sum of the atomic numbers as is, which would result in the original isotope being $^{14}_7$N.
d) To find the missing value, we would add one neutron to the mass number of $^{272}_{111}$Rg, and leave the atomic number as is. We would then subtract 64 from the mass number and 28 from the atomic number to get and original isotope equaling $^{209}_{83}$Bi.
