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Question

When 3Li7 nuclei are bombarded by protons, and the resultant nuclei are 4Be8 , the emitted particles will be.

A
alpha particles
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B
beta particles
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C
gamma photons
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D
neutrons
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Solution

The correct option is C gamma photons
3Li7+11P4Be8+00γ(energy)
As, the emitted particles is only energy and gamma photons radiation is nothing but energy so, option C is correct.

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