The particle $X$ in the following nuclear reaction is $_{5}^{B} 10$ + $_{2}^{H e} 4$ $⟶$ $_{7}^{N} 13$ + $X$

P

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a

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e

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n

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Solution

The correct option is D $n$
Since the atomic number remains same (i.e $5 + 2 = 7$ ), but mass number is changed, it is a neutron.

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Similar questions

_{5} B^{10} +_{2} H e^{4} \to_{7} N^{13} +_{0} n^{1} \to_{6} C^{13} +_{+ 1} e^{0}

n \to p + e^{-} + X

X

_{7}^{13} N

⟶

_{6}^{13} C +

_{1}^{0} e

X

$_{5} B^{10}$ + $_{2} H e^{4}$ $\to$ X + $_{0} n^{1}$

'X' in the above reaction is

_{2} H e^{4} +_{7} N^{14} \to_{1} H^{1} + X

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