Question
In the given reaction, find value of Q value. 6C13 → 6C12 + 0n1 + (Q-value) Given : mass of 6C13 ⇒ x mass of 6C12 ⇒ y mass of 0n1 ⇒ z
Options :
(y + x – z) C2
(y + z – x) C2
(y + z + x) C2
(z + x – y) C2
Answer :
(y + z – x) C2
Solution :
⇒ △m = (y + z - x)
Q - value = △mC2
= (y + z - x)C2
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