Question

When Uranium is bombarded with neutrons, it undergoes fission. The fission reaction can be written as
${{}_{92}U}^{235}+{{}_{0}n}^1⟶{{}_{56}Ba}^{141}+{{}_{36}Kr}^{92}+3x+Q\left(energy\right)$
Where three particles named $x$ are produced and energy $Q$ is released. What is the name of the particle $x$?

• A. electron
• B. $\alpha$- particle
• C. neutron
• D. neutrino

Answer 1

Answer: (C)

given reaction,

${}_{92}{U}^{235}+{}_0{n}^1\to {}_{56}B{a}^{141}+{}_{36}K{r}^{92}+3x+Q\left(energy\right)$

$\sum$ Atomic number on L.H.S. $=92+0=92$

$\sum$ Atomic number on R.H.S. $=56+36=92$

$\therefore$ number of protons on L.H.S. are equal to the number of protons on R.H.S. hence x not contains protons in it.

$\sum$ Atomic mass number on L.H.S. $=235+1=236$

$\sum$ Atomic mass number on R.H.S $=141+92+3x=233+3x$

L.H.S$=$R.H.S

$233+3x=236$

$x=1$

Mass number$=1$, no. of proton $=0$

therefor particle will be Neutron.