Question

The missing particle in the reaction ${}_{92}^{235}U{+}_0^{1}n{\to }_{56}B{a}^{146}+\cdots +3_0^{1}n$ is:

• A. ${}_{32}^{87}Ge$
• B. ${}_{35}^{89}Br$
• C. ${}_{36}^{87}Kr$
• D. ${}_{35}^{86}Br$

Answer 1

The correct option is C ${}_{36}^{87}Kr$

It is a neutron-induced fission reaction.

The Sum of atomic masses on the reactant side should be equal to the sum of atomic masses on the product side.

$\therefore$ mass of missing particle $+146+(3\times 1)=235+1$

$\Rightarrow$ mass of missing particle $+149=236$

$\Rightarrow$ mass of missing particle $=236-149=87$

The Sum of atomic numbers on the reactant side should be equal to the sum of atomic numbers on the product side.

$\therefore$ atomic number of missing particle $+56+(3\times 0)=92+0$

$\Rightarrow$ atomic number of missing particle $+56=92$

$\Rightarrow$ atomic number of missing particle $=92-56=36$

Hence, option $C$ is correct.