Answer:
Let \[_{Z}^{A}\] P be the bombarding particle. Then \[_{3}^{7}Li+_{Z}^{A}P\to _{2}^{4}He+_{2}^{4}He\] Using the laws of conservation of mass and energy, we get \[A+7=4+4\] or \[A=1\] \[Z+3=2+2\] or \[Z=1\] Thus, the bombarding particle is a proton \[(_{1}^{1}H)\].
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