When a uranium isotope 235/92 U is bombarded with a neutron, it generates 89/36 Kr, three neutrons and: 91/40 Zr 101/36 Kr 103/36 Kr 144/56 Ba

Key ideas you must know

1. Conservation laws in nuclear reactions

   • Mass number $A$ (total protons + neutrons) is conserved.
   • Atomic number $Z$ (total protons, i.e., charge) is conserved.

2. Typical fission of $^{235}_{92}\text U$
   When $^{235}_{92}\text U$ absorbs a neutron it becomes an excited $^{236}_{92}\text U$ which then splits. Many different fragment pairs are possible. In this question, one of the fragments is already given: $^{89}_{36}\text{Kr}$. Three free neutrons are also given. We must find the other fragment.

How a student should proceed

1. Write the skeleton reaction
   $^{235}_{92}\text U + ^1_0\text n \;\longrightarrow\; ^{89}_{36}\text{Kr} + ^A_Z\text X + 3\, ^1_0\text n$

2. Balance mass numbers
   Initial: $235 + 1 = 236$
   Final: $89 + A + 3(1)$
   $236 = 89 + A + 3 \;\Rightarrow\; A = 236 - 89 - 3 = 144$

3. Balance atomic numbers
   Initial: $92$
   Final: $36 + Z + 3(0)$
   $92 = 36 + Z \;\Rightarrow\; Z = 56$

4. Identify the element with $Z = 56$
   From the periodic table, $Z = 56$ is Barium (Ba). Therefore the missing fragment is $^{144}_{56}\text{Ba}$.

Hence, option 144/56 Ba is correct.

Imagine breaking a big LEGO block

Think of the heavy uranium-235 nucleus as a huge LEGO block. When a tiny LEGO piece (a neutron) flies in and hits it, the big block snaps into two smaller blocks plus a few little pieces (extra neutrons).

But LEGO never disappears! The total number of studs (mass number) and the color code (atomic number) before and after the break must stay the same. So we just do ordinary counting to find out which second block is missing.

Step-by-step working

1. Write the reaction skeleton
   $^{235}_{92}\text U + ^1_0\text n \;\to\; ^{89}_{36}\text{Kr} + \;^{A}_{Z}\text X + 3\, ^1_0\text n$

2. Balance mass number

   $235 + 1 = 89 + A + 3 \times 1$ $236 = 89 + A + 3$ $A = 236 - 89 - 3 = 144$

3. Balance atomic number

   $92 = 36 + Z + 3 \times 0$ $Z = 92 - 36 = 56$

4. Identify the element
   $Z = 56$ corresponds to Barium (Ba).

5. Final balanced reaction

   $^{235}_{92}\text U + ^1_0\text n \;\longrightarrow\; ^{89}_{36}\text{Kr} + ^{144}_{56}\text{Ba} + 3\, ^1_0\text n + Q$

6. Answer: $^{144}_{56}\text{Ba}$

Hence, the correct option is 144/56 Ba.