bombarding plutonium - 239 with an alpha particle creates an isotope of curium. balance the…
bombarding plutonium - 239 with an alpha particle creates an isotope of curium. balance the equation.\n$_{94}^{239}pu+_{2}^{4}he\\rightarrow_{b}^{a}cm + _{d}^{c}n$\nidentify the values for the unknown letters.\na is \nb is \nc is \nd is
Answer
Explanation:
Step1: Balance mass - number
The sum of mass - numbers on the left - hand side of the nuclear equation must equal the sum of mass - numbers on the right - hand side. The mass - number of $^{239}{94}Pu$ is 239 and the mass - number of $^{4}{2}He$ is 4. So, $239 + 4=a + c$. In a typical nuclear reaction where a neutron ($^{1}_{0}n$) is produced, we assume $c = 1$. Then $a=239 + 4-1=242$.
Step2: Balance atomic - number
The sum of atomic - numbers on the left - hand side of the nuclear equation must equal the sum of atomic - numbers on the right - hand side. The atomic - number of $^{239}{94}Pu$ is 94 and the atomic - number of $^{4}{2}He$ is 2. So, $94+2 = b + d$. Since a neutron has an atomic - number of 0 ($d = 0$), then $b=94 + 2=96$.
Answer:
a is 242 b is 96 c is 1 d is 0