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Physics
When a 4Be9 is bombarded with α- particle, one of the products of nuclear transmutations is 6C12. If other is
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Q. When a $_4Be^{9}$ is bombarded with $\alpha$- particle, one of the products of nuclear transmutations is $_6C^{12}$. If other is
Atoms
A
$_{-1}e^{0}$
8%
B
$_1H^{1}$
24%
C
$_1H^{2}$
28%
D
$_0n^{1}$
40%
Solution:
$_4Be^9 + _2He^4 \rightarrow _6C^{12} + _0n^1$