Q.The missing particle from the nuclear reaction is
27-13-Al + 4-2-He → ? + 1-0-n
a. 30-15-P
b. 32-16-S
c. 14-10-Ne
d. 14-Si

Step 1. In 27-13-Al + 4-2-He → ? + 1-0-n, conserve total mass number: 27 + 4 = 31 on the left, so the missing particle's mass number = 31 - 1 (for the ejected neutron) = 30.

Step 2. Conserve total atomic number: 13 + 2 = 15 on the left, and the ejected neutron carries atomic number 0, so the missing particle's atomic number = 15 - 0 = 15.

Step 3. Atomic number 15 is phosphorus, so the missing particle is 30-15-P -- matching option (a).

Step 4. This is in fact a real, well-known (α,n) transmutation of aluminium-27 into phosphorus-30 (historically significant as one of the first artificially produced radioactive isotopes), directly analogous to the Al-27(α,n) reaction described in this chapter's own 'Try this' box.

Balance total mass number and total atomic number across the nuclear equation to identify the missing product.

• Forgetting to subtract the ejected neutron's mass number (1) when solving for the missing particle's mass number.
• Picking an option with the right mass number but wrong atomic number, or vice versa, instead of checking both conservation laws together.