When Uranium is bombarded with neutrons, it undergoes fission. The fission reaction can be written as :
$$_{92}U^{235} + {}_0n^1 \rightarrow {}_{56}Ba^{141} + {}_{36}Kr^{92} + 3x + Q$$(energy) where three particles named x are produced and energy Q is released. What is the name of the particle x?

We are given the fission reaction of Uranium-235 when bombarded with a neutron:

$$_{92}U^{235} + {}_0n^1 \rightarrow {}_{56}Ba^{141} + {}_{36}Kr^{92} + 3x + Q$$

Here, Q represents the energy released, and x is the unknown particle we need to identify. To find x, we must apply the conservation laws of mass number and atomic number. These laws state that the total mass number (sum of superscripts) and total atomic number (sum of subscripts) must be equal on both sides of the reaction.

First, calculate the total mass number on the left side (reactants):

Uranium-235 has a mass number of 235, and the neutron ($$ {}_0n^1 $$) has a mass number of 1. So, total mass number left = 235 + 1 = 236.

Total atomic number on the left side: Uranium has atomic number 92, and the neutron has atomic number 0. So, total atomic number left = 92 + 0 = 92.

Now, for the right side (products): Barium-141 has mass number 141 and atomic number 56. Krypton-92 has mass number 92 and atomic number 36. The three particles x are identical, so let each have mass number A and atomic number Z. Therefore, the total contribution from 3x is mass number 3A and atomic number 3Z. Energy Q has no mass or charge, so it doesn't affect these numbers.

Total mass number right = mass number of Ba + mass number of Kr + mass number from 3x = 141 + 92 + 3A.

Total atomic number right = atomic number of Ba + atomic number of Kr + atomic number from 3x = 56 + 36 + 3Z.

Apply conservation of mass number:

Left side total mass number = Right side total mass number

236 = 141 + 92 + 3A

236 = 233 + 3A

Subtract 233 from both sides:

236 - 233 = 3A

3 = 3A

Divide both sides by 3:

A = 1

Apply conservation of atomic number:

Left side total atomic number = Right side total atomic number

92 = 56 + 36 + 3Z

92 = 92 + 3Z

Subtract 92 from both sides:

92 - 92 = 3Z

0 = 3Z

Divide both sides by 3:

Z = 0

So each particle x has a mass number of 1 and an atomic number of 0. This matches the properties of a neutron ($$ {}_0n^1 $$), which has no charge (atomic number 0) and a mass number of 1.

Now, check the options:

• Electron: Mass number is approximately 0 (very light), atomic number is -1 (charge -1). Does not match.
• Alpha-particle: Mass number 4 (helium nucleus), atomic number 2. Does not match.
• Neutron: Mass number 1, atomic number 0. Matches.
• Neutrino: Mass number approximately 0, atomic number 0. Does not match (mass number not 1).

Therefore, the particle x is a neutron. In nuclear fission, such as this reaction, multiple neutrons are typically emitted.

Hence, the correct answer is Option C.