MCQ
In the nuclear reaction: $X(n,\,\alpha ){\,_3}L{i^7}$ the term $X$ will be
- ✓$_5{B^{10}}$
- B$_5{B^9}$
- C$_5{B^{11}}$
- D$_2H{e^4}$
✓
Answer
Correct option: A.
$_5{B^{10}}$
a
(a) $X(n,\;\alpha )\;_3^7Li{ \Rightarrow _Z}{X^A}{ + _0}{n^1} \to 3L{i^7}{ + _2}H{e^4}$
(a) $X(n,\;\alpha )\;_3^7Li{ \Rightarrow _Z}{X^A}{ + _0}{n^1} \to 3L{i^7}{ + _2}H{e^4}$
$Z = 3 + 2 = 5$ and $A= 7+4 -1 =10$
${\therefore _5}{X^{10}}{ = _5}{B^{10}}$
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