2713Al + 42He  →   3015P  +  _____A. 110nB. 210nC. 310nD. 410n

This is a nuclear reaction and we can solve for the missing particle by conserving the number of protons and neutrons before and after the reaction.

Before the reaction, we have:

• Number of protons = 27 (from Al) + 2 (from He) = 29
• Number of nucleons (protons + neutrons) = 13 (from Al) + 4 (from He) = 17

After the reaction, we have:

• Number of protons = 30 (from P)
• Number of nucleons = 15 (from P) + n (from the missing particle)

By conserving the number of protons, we have 29 = 30 + p (where p is the number of protons in the missing particle). This gives p = -1, which is not possible because the number of protons cannot be negative.

By conserving the number of nucleons, we have 17 = 15 + n (where n is the number of nucleons in the missing particle). This gives n = 2.

Therefore, the missing particle is 2 neutrons, so the correct answer is B. 2^10n.