Question

${{}_{92}U}^{235}$ nucleus absorbs a neutron and disintegrates into ${{}_{54}Xe}^{139},{{}_{38}Sr}^{94}$ and $X$. What will be the product $X$?

• A. 2 neutrons
• B. 3 neutrons
• C. $\alpha$-particle
• D. $\beta$-particle

Answer 1

Answer: (B)

3 neutrons

${{}_{92}U}^{235}{+}_0{n}^1\to {{}_{54}Xe}^{139}+{{}_{38}Sr}^{94}+m_A{X}^B$

$92+0=54+38+A$

$A=0$

$235+1=139+94+mB$

$mB=3$
Thus, the product $X$ will be 3 neutrons.
${{}_{92}U}^{235}{+}_0{n}^1\to {{}_{54}Xe}^{139}+{{}_{38}Sr}^{94}+3_0{n}^1$