Question

Which nuclide is formed as a result of this fission reaction?
${}_{92}^{235}U{+}_0^{1}n\to 3_0^{1}n{+}_{56}^{139}Ba+\_\_\_\_\_\_\_\_\_\_$

• A. ${}_{35}^{96}Br$
• B. ${}_{36}^{96}Kr$
• C. ${}_{37}^{94}Rb$
• D. ${}_{36}^{94}Kr$
• E. ${}_{38}^{90}Sr$

Answer 1

The correct option is D ${}_{36}^{94}Kr$

${}_{92}^{235}U{+}_0^{1}n\to 3_0^{1}n{+}_{56}^{139}Ba{+}_Z^{A}X$

$\bullet$ Finding Z

92 = 56+Z

Z = 36

$\Rightarrow$ X is Krypton(Kr)

$\bullet$ Finding A

235 + 1 = 3 + 139 + A

A = 94

$\therefore$ ${}_Z^{A}X\equiv {}_{36}^{94}Kr$