Final answer:
The balanced nuclear equation for the alpha decay of berkelium-248 is represented as 248/97 Bk -> 244/95 Am + 4/2 He, indicating that berkelium-248 decays into americium-244 and an alpha particle.
Step-by-step explanation:
To write the balanced nuclear equation for the alpha decay of berkelium-248, we need to represent both the initial radioactive isotope and the products of decay, which include an alpha particle and the daughter isotope. An alpha particle consists of 2 protons and 2 neutrons, which means its symbol is 4He2. Because the atomic number of berkelium is 97, and its mass number is 248, the initial isotope is represented as 248Bk97.
During alpha decay, the parent nucleus emits an alpha particle (helium nucleus), resulting in the loss of 2 protons and 2 neutrons. Thus, the atomic number will decrease by 2 and the mass number will decrease by 4. The daughter isotope will have an atomic number of 95 (97 - 2) and a mass number of 244 (248 - 4). The element with atomic number 95 is americium (Am).
The balanced nuclear equation for this process is:
248Bk97 → 244Am95 + 4He2.