When 3​ Li 7 nuclei are bombarded by protons and the resultant nuclei are 4​ Be 8 , the emitted particles will be:

The reaction can be written as follows:

3Li7 + 1H1 -> 4Be8 + X

Where X is the particle emitted. To find X, we need to balance the reaction.

On the left side of the reaction, we have a total atomic number of 3 (from Li) + 1 (from H) = 4. On the right side, we have an atomic number of 4 (from Be). So, the atomic number is balanced.

Next, we balance the mass number. On the left side, we have a total mass number of 7 (from Li) + 1 (from H) = 8. On the right side, we have a mass number of 8 (from Be). So, the mass number is also balanced.

Since both the atomic number and mass number are balanced, no particle is emitted in this reaction. Therefore, X is nothing.