When $X ightarrow ^{14}_{7} ext{N} + 2\beta^-$, then the number of neutrons in $X$ will be:

Question

When $X ightarrow ^{14}_{7}	ext{N} + 2\beta^-$, then the number of neutrons in $X$ will be:

Accepted Answer

In a nuclear reaction, both the total mass number ($A$) and the total atomic number ($Z$) are conserved. The given reaction is $X ightarrow ^{14}_{7}	ext{N} + 2\beta^-$. A $\beta^-$ particle is an electron, represented as $^0_{-1}e$.

1.  **Nuclear Equation:** Let $X$ be represented as $^A_Z X$. The equation is: 
    $^A_Z X ightarrow ^{14}_{7}	ext{N} + 2(^{0}_{-1}e)$

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

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

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

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

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