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NEET PHYSICSNUCLEIEasy

Question

When X714N+2βX \rightarrow ^{14}_{7}\text{N} + 2\beta^-, then the number of neutrons in XX will be:

A

3

B

5

C

7

D

9

Step-by-Step Solution

In a nuclear reaction, both the total mass number (AA) and the total atomic number (ZZ) are conserved. The given reaction is X714N+2βX \rightarrow ^{14}_{7}\text{N} + 2\beta^-. A β\beta^- particle is an electron, represented as 10e^0_{-1}e.

  1. Nuclear Equation: Let XX be represented as ZAX^A_Z X. The equation is: ZAX714N+2(10e)^A_Z X \rightarrow ^{14}_{7}\text{N} + 2(^{0}_{-1}e)

  2. Conservation of Mass Number (AA): A=14+2(0)=14A = 14 + 2(0) = 14

  3. Conservation of Atomic Number (ZZ): Z=7+2(1)=72=5Z = 7 + 2(-1) = 7 - 2 = 5

  4. Number of Neutrons (NN): The number of neutrons is calculated as N=AZN = A - Z. N=145=9N = 14 - 5 = 9.

Therefore, the number of neutrons in the parent nucleus XX is 9.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from NUCLEI. 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.

PHYSICSNUCLEIrightarrownumberneutrons

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