Study the diagram given below and answer the questions that follow :

(a) Nuclear fission.

(b) X = 236

(c) As, the mass number and atomic number is conserved in the reaction then

From conservation of mass number

Let, mass number of Y be y.

mass number of neutron + mass number of $_{92}^{235}\text{U}$ = 3 x mass number of Y + mass number of $_{36}^{89}\text{Kr}$ + mass number of $_{56}^{144}\text{Ba}$

$$\begin{aligned}1 + 235 = 3 \times \text y + 89 + 144 \\[1em] ⇒ 236 = 3 \text y + 233 \\[1em] ⇒ 3 \text y = 236 - 233 = 3 \\[1em] ⇒ \text y = \frac{3}{3}=1 \\[1em]\end{aligned}$$

So, mass number of Y is 3.

From conservation of atomic number

Let, atomic number of Y be z.

atomic number of neutron + atomic number of $_{92}^{235}\text{U}$ = atomic number of Y + atomic number of $_{36}^{89}\text{Kr}$ + atomic number of $_{56}^{144}\text{Ba}$

$$\begin{aligned}⇒ 92 + 0 = \text z + 36 + 56 \\[1em] ⇒ 92 = \text z + 92 \\[1em] ⇒ \text z = 92 - 92 = 0 \\[1em]\end{aligned}$$

As mass number of Y is 1 and it's atomic number is zero which is a property of a neutron.

Hence, Y is a neutron.