Problem 64 - A Complete Resource Book in Chemistry for JEE Main

Question

\({ }_{92} \mathrm{U}^{235}\) nucleus absorbs a neutron and disintegrates into \({ }_{54} \mathrm{Xe}^{139},{ }_{38} \mathrm{Sr}^{94}\) and \(\mathrm{x}\). What will be the product \(\mathrm{x}\) ? (a) 3 neturons (b) 2 neturons (c) \(\alpha\) particle (d) \(\beta\) particle

Step-by-Step Solution

Answer

The product \( x \) is 3 neutrons, option (a).

Step 1: Write the Nuclear Reaction Equation

The uranium-235 nucleus absorbs a neutron, thus the initial reaction is: \[ { }_{92} \mathrm{U}^{235} + { }_{0} \mathrm{n}^1 \rightarrow { }_{54} \mathrm{Xe}^{139} + { }_{38} \mathrm{Sr}^{94} + \mathrm{x} \] We need to find out what \( \mathrm{x} \) is by balancing the nuclear equation.

Step 2: Check Conservation of Mass Number

The mass number before the reaction is \( 235 + 1 = 236 \). After the reaction, the mass is the sum of \( 139 + 94 + A_x \) (where \( A_x \) is the mass number of \( x \)). Thus, we have: \[ 236 = 139 + 94 + A_x \] Solving for \( A_x \), we get \( A_x = 3 \).

Step 3: Check Conservation of Atomic Number

On the left, the atomic number is \( 92 + 0 = 92 \). After the reaction, it should be \( 54 + 38 + Z_x \), where \( Z_x \) is the atomic number of \( x \). Thus, \[ 92 = 54 + 38 + Z_x \] Solving for \( Z_x \), we get \( Z_x = 0 \).

Step 4: Determine the Identity of \( x \)

With a mass number of \( 3 \) and an atomic number of \( 0 \), the product \( x \) must be 3 neutrons. Thus, \( x \) is \( 3 \) neutrons.

Key Concepts

Mass Number Conservation

In nuclear reactions, the mass number must be conserved. This means that the total mass number of all particles before and after the reaction remains unchanged. For our example, we start with the uranium-235 nucleus absorbing a neutron. The initial mass number is calculated from the uranium and neutron:

• Uranium-235: 235
• Neutron: 1

Therefore, before the reaction, the total mass number is 236. After the reaction, we must account for the mass numbers of xenon-139 and strontium-94, along with the unknown product, x. So: \[ 236 = 139 + 94 + A_x \] Solving this equation gives us the mass number of x, which is 3. Maintaining this balance is crucial to identifying the unknown product in any nuclear reaction.

Atomic Number Conservation

The atomic number reflects the number of protons in an atomic nucleus and must be conserved in nuclear reactions. Conservation of atomic number means that the total number of protons before and after the reaction should be the same.
For the uranium-235 absorbing a neutron, the atomic numbers initially are:

• Uranium-235: 92
• Neutron: 0

This totals to an atomic number of 92 initially. After the reaction, we sum the atomic numbers of xenon-139 (54) and strontium-94 (38), along with the unknown product: \[ 92 = 54 + 38 + Z_x \] Solving here shows that the atomic number of x is 0. This helps us deduce that x cannot be a charged particle but must be neutral, like a neutron.

Neutron Emission

Neutron emission refers to the release of neutrons from a nucleus during a nuclear reaction or decay. Neutrons, being neutral particles without charge, affect only the total mass number but not the atomic number.
In the exercise, after identifying the conservation of mass and atomic numbers, the solution reveals that the extra mass is composed of 3 neutrons.

• Neutrons have a mass number of 1 each.
• They do not alter the atomic number.

Hence, a mass number for x as 3 and an atomic number of 0 correctly indicates that three neutrons were emitted. Understanding neutron emission is key to solving this problem and highlights why they do not affect the chemical identity of the nucleus.