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NEET CHEMISTRYChemical KineticsMedium

Question

92235U^{235}_{92}U nucleus absorbs a neutron and disintegrates into 54139Xe^{139}_{54}Xe, 3894Sr^{94}_{38}Sr and xx. The product xx is:

A

3 neutrons

B

2 neutrons

C

α\alpha - particle

D

β\beta - particle

Step-by-Step Solution

In a nuclear reaction, both the mass number (A) and the atomic number (Z) are conserved. The given reaction can be written as:

92235U+01n54139Xe+3894Sr+x^{235}_{92}U + ^{1}_{0}n \rightarrow ^{139}_{54}Xe + ^{94}_{38}Sr + x

Applying the law of conservation of mass number (superscripts): 235+1=139+94+Ax235 + 1 = 139 + 94 + A_x 236=233+Ax    Ax=3236 = 233 + A_x \implies A_x = 3

Applying the law of conservation of atomic number (subscripts): 92+0=54+38+Zx92 + 0 = 54 + 38 + Z_x 92=92+Zx    Zx=092 = 92 + Z_x \implies Z_x = 0

The product xx has a total mass number of 3 and an atomic number of 0. This corresponds to 3 neutrons (3×01n3 \times ^{1}_{0}n).

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Chemical Kinetics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYChemical Kineticsnucleusabsorbsneutrondisintegratesproduct

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