Problem 29 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Complete and balance the following reactions. \({ }_{92} \mathrm{Th}^{234} \longrightarrow \ldots \ldots \ldots . .+7{ }_{2} \mathrm{He}^{4}+6_{-1}^{(i)} \beta^{0}\) (ii) \({ }_{92} \mathrm{U}^{235}+{ }_{0} \mathrm{n}^{1} \longrightarrow \ldots \ldots \ldots .+{ }_{52} \mathrm{Te}^{137}+{ }_{40} \mathrm{Zr}^{92}\) (iii) \({ }_{34} \mathrm{Se}^{86} \longrightarrow 2_{-1} \mathrm{e}^{0}+\ldots \ldots \ldots \ldots .\)

Step-by-Step Solution

Answer

(i) Po-206; (ii) 7 neutrons; (iii) Kr-86.

Step 1: Understand the types of reactions

For part (i), the reaction involves a thorium-234 atom releasing alpha particles and beta particles. In part (ii), uranium-235 undergoes nuclear fission after capturing a neutron, resulting in the production of two nuclei. In part (iii), selenium-86 undergoes a nuclear transformation with electron emission.

Step 2: Find missing products in Reaction (i)

For the equation \( {}_{92}^{234} \text{Th} \rightarrow x \text{ + } 7_{2}^{4} \text{He} + 6_{-1}^{0} \beta \), calculate the mass number and atomic number of the missing element "x." To balance the reaction: \[ A_x = 234 - (7 \times 4) = 206 \] \[ Z_x = 92 - (7 \times 2) + 6 = 84 \] Hence, the missing element is \( {}_{84}^{206} \text{Po} \) (Polonium).

Step 3: Balance Reaction (ii)

The equation is \( {}_{92}^{235} \text{U} + {}_{0}^{1} \text{n} \rightarrow x + {}_{52}^{137} \text{Te} + {}_{40}^{92} \text{Zr} \). To find "x," calculate the remaining mass and atomic number: \[ A_x = 236 - (137 + 92) = 7 \] \[ Z_x = 92 - (52 + 40) = 0 \] This indicates there are 7 neutrons unaccounted, so "x" is \( 7_{0}^{1} \text{n} \).

Step 4: Identify missing element in Reaction (iii)

The equation is \( {}_{34}^{86} \text{Se} \rightarrow 2_{-1}^{0} e + x \). Calculate the mass and atomic number of the missing element "x": \[ A_x = 86 \; (since \; electrons \; have \; negligible \; mass) \] \[ Z_x = 34 + 2 = 36 \] The resulting element is \( {}_{36}^{86} \text{Kr} \), which means the product is Krypton.

Key Concepts

Alpha Decay

Alpha decay is a process in nuclear physics where an unstable nucleus releases an alpha particle to become more stable. An alpha particle consists of 2 protons and 2 neutrons, identical to a helium-4 nucleus. This decay typically occurs in heavy elements where the repulsion between protons becomes too strong.
When a nucleus undergoes alpha decay, its mass number decreases by 4, and its atomic number decreases by 2. This transforms the original element into a new one by moving it two places back in the periodic table. For example, the alpha decay from thorium to polonium involves the release of 7 alpha particles in the given problem.

• The mass number of polonium is calculated as: 234 - (7×4) = 206.
• The atomic number is: 92 - (7×2) + 6 = 84.

Thus, alpha decay plays a critical role in reducing the size of heavy nuclei and altering their identities in the periodic table.

Nuclear Fission

Nuclear fission is a reaction in which a nucleus splits into smaller nuclei, releasing energy in the process. It commonly occurs when a heavy nucleus, like Uranium-235, captures a neutron, making it unstable and causing it to break apart.
In the exercise, when uranium-235 undergoes fission, it captures one neutron (making a total mass number of 236) and produces two smaller nuclei: Tellurium-137 and Zirconium-92.

• The neutron initiates the reaction by destabilizing the uranium nucleus.
• The remaining mass and atomic numbers indicate there are 7 additional neutrons released: \[ A_x = 236 - (137 + 92) = 7, \ Z_x = 92 - (52 + 40) = 0. \]

The energy released from fission reactions is harnessed in nuclear power plants to generate electricity, showcasing its importance in energy production.

Beta Decay

Beta decay is a process by which an unstable atomic nucleus changes composition by emitting a beta particle. A beta particle can either be an electron or a positron, which results from the conversion of a neutron to a proton, or vice versa.
In beta decay, the original nucleus stays almost the same in mass but changes its atomic number. This occurs because a neutron is transformed into a proton, or a proton into a neutron, altering the nucleus's charge.
For example, in the problem, selenium-86 undergoes beta decay, resulting in:

• Mass number: Remains 86, as electrons carry negligible mass.
• Atomic number: Changes from 34 to 36, adding 2 protons, because of 2 emitted electrons.

The product is Krypton-86, illustrating how beta decay allows a nucleus to adjust its proton-to-neutron ratio for greater stability while changing its identity on the periodic table.