Question

When a ${}_{92}^{235}\text{U}$ isotope is bombarded with a neutron it generates ${}_{51}^{133}\text{Sb}$, four neutrons and

• A. ${}_{38}^{94}\text{Sr}$
• B. ${}_{54}^{140}\text{Xe}$
• C. ${}_{41}^{99}\text{Nb}$
• D. ${}_{36}^{89}\text{Kr}$

Answer 1

The correct option is C ${}_{41}^{99}\text{Nb}$

${}_{92}^{235}\text{U}{+}_0^1\text{n}{\to }_{51}^{133}\text{Sb}{+}_Z^A\text{X}+4{}_0^1\text{n}$
Total number of nucleons remains constant in the nuclear reaction.

$\therefore 92=51+Z+0\Rightarrow Z=92–51=41$

$235+1=133+A+4$
$236=137+A$
$\therefore A=236–137=99$
${\therefore }_Z^A\text{X}={}_{41}^{99}\text{Nb}$