In given nuclear reaction, \(^9_4Be\) + α → \(^{12}_6C\) + X, pa

1.	In given nuclear reaction, \(^9_4Be\) + α → \(^{12}_6C\) + X, particle X is(a) \(^4_2He\)(b) \(^0_{-1}e\)(c) \(^1_1H\)(d) \(^1_0n\)
Answer» Answer is : (d) \(^1_0n\) Following conservation of mass number and atomic number, we have \(^9_4Be\) + \(^4_2He\) → \(^{12}_6C\) + \(^1_0n\) So, particle \(^1_0n\) is a neutron.

In given nuclear reaction, \(^9_4Be\) + α → \(^{12}_6C\) + X, particle X is(a) \(^4_2He\)(b) \(^0_{-1}e\)(c) \(^1_1H\)(d) \(^1_0n\)

Answer»

Answer is : (d) \(^1_0n\)

Following conservation of mass number and atomic number, we have

\(^9_4Be\) + \(^4_2He\) → \(^{12}_6C\) + \(^1_0n\)

So, particle \(^1_0n\) is a neutron.

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In given nuclear reaction, \(^9_4Be\) + α → \(^{12}_6C\) + X, particle X is(a) \(^4_2He\)(b) \(^0_{-1}e\)(c) \(^1_1H\)(d) \(^1_0n\)

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