(B) The given nuclear reaction is: $_{5}B^{10} + _{2}He^{4} \rightarrow _{b}^{a}X + _{0}n^{1}$ By balancing the atomic numbers (sum of protons): $5 +

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(B) The given nuclear reaction is: $_{5}B^{10} + _{2}He^{4} \rightarrow _{b}^{a}X + _{0}n^{1}$ By balancing the atomic numbers (sum of protons): $5 + 2 = b + 0$ $b = 7$ By balancing the mass numbers (sum of protons and neutrons): $10 + 4 = a + 1$ $14 = a + 1$ $a = 13$ Therefore, the product $X$ is $_{7}N^{13}$.

Class 12 Chemistry Nuclear Chemistry Artificial transmutation

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$_{5}B^{10} + _{2}He^{4} \rightarrow X + _{0}n^{1}$
In the above nuclear reaction '$X$' will be

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A
$_{7}N^{14}$
B
$_{7}N^{13}$
C
$_{6}C^{12}$
D
$_{7}N^{15}$

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$_{5}B^{10} + _{2}He^{4} \rightarrow X + _{0}n^{1}$ In the above nuclear reaction '$X$' will be

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$_{5}B^{10} + _{2}He^{4} \rightarrow X + _{0}n^{1}$
In the above nuclear reaction '$X$' will be