Question

In the nuclear reaction,

${}_{92}{U}^{235}+0_{N}1{\to }_{5}6B{a}^{141}+K{r}^{92}+3X+200MeV$

$X$ represents:

• A. Proton
• B. Neutron
• C. Electron
• D. Alpha particle

Answer 1

Answer: (B)

Neutron

In following nuclear reaction :

${}_{92}{U}^{235}{+}_0{n}^1{\to }_{56}B{a}^{141}{+}_{36}{K}^{89}+3X+200MeV$

We have to find the value of $X$

So Atomic Number of L.H.S=R.H.S

$92=56+36(=92)$

Mass number of L.H.S= R.H.S

$L.H.S\to$ $235+1=236$

$R.H.S\to$ $141+92=233$

So Atomic mass of $3$ should be added in order to complete this reaction.

We know that Neutron has mass number=$1$ and $0$ charge so in this reaction $30_n^1$ must be eliminated.

Hence $X=Neutron$

Option B is correct.