Question

If ${}_{92}{U}^{235}$ nucleus absorbs a neutron and disintegrates in ${}_{54}X{e}^{139},{}_{38}S{r}^{94}$ and X, then what will be the product X?

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

Answer 1

The correct option is D 3-neutrons

The required reaction is:

${}_{92}{U}^{235}{+}_0{n}^1{\to }_{54}X{e}^{139}{+}_{38}S{r}^{94}+3_0{n}^1$

It results in 3 neutrons.

Hence, option D is correct.