Question

Write the complete nuclear equation for the bombardent of a 27 ^Al atom with an α particle to yield 30^ P. Show the atomic number and mass number for each species in the equation.

Answer 1

The complete nuclear equation for the bombardment of a 27 Al atom with an α particle to yield 30 P is 27 Al + 2 He → 30 P.

Step-by-step explanation:

When bombarding an element with an alpha particle (symbolized as α), the resulting element will have an atomic number that is 2 greater than the target nucleus. In this case, aluminum (Al) has an atomic number of 13, so the resulting element will be phosphorus (P) with an atomic number of 15.

Additionally, the mass number of the resulting element must be 3 greater than the target nucleus to conserve mass. Since the mass number of the target nucleus is 27, the mass number of the resulting element will be 30.

Therefore, the complete nuclear equation for the bombardment of a 27 Al atom with an α particle to yield 30 P is:

27 Al + 2 He → 30 P

Answer 2

When an α particle bombards an aluminum-27 atom, it yields phosphorus-30. The balanced nuclear equation is 27^Al + 2^He → 30^P + 1^H.

Step-by-step explanation:

When an alpha particle (α particle) bombards an aluminum-27 (27^Al) atom, it yields phosphorus-30 (30^P).

The atomic number of the target nucleus, aluminum, is 13, and after the reaction, it increases to 15, which is the atomic number of phosphorus.

The mass number of the target is 27, and the mass number of the product, phosphorus-30, is 30, which indicates that a neutron is released during the reaction.

Therefore the complete balanced nuclear equation for this reaction is: 27^Al + 2^He → 30^P + 1^H