An element of P and atomic Q emits an alpha particle from its nucleus thet resultant numbers of the new element formed are respectively
an alpha particle is essentially a helium-4 nucleus
... two protons and two neutrons
the new element would have an atomic number that is lower by two
... and a mass number that is lower by four
In order to determine the new elements formed when an element of P and atomic Q emits an alpha particle, we need to understand the process of alpha decay and how new elements are formed.
Alpha decay is a type of radioactive decay in which an atomic nucleus emits an alpha particle. An alpha particle consists of two protons and two neutrons, which are essentially the same as a helium nucleus. As a result, when an element undergoes alpha decay, its atomic number decreases by 2 and its mass number decreases by 4.
To determine the new elements formed, we need to know the atomic number (Z) and atomic mass number (A) of the original elements in P and Q. Let's assume the original element in P has an atomic number of Zp and atomic mass number of Ap, and the original element in Q has an atomic number of Zq and atomic mass number of Aq.
When an alpha particle is emitted, the new element formed will have an atomic number of Zp-2 and an atomic mass number of Ap-4. Similarly, for element Q, the new element formed will have an atomic number of Zq-2 and an atomic mass number of Aq-4.
So, the resultant numbers of the new elements formed are:
Element from P: (Zp-2, Ap-4)
Element from Q: (Zq-2, Aq-4)
To determine the specific elements formed, we would need to know the initial atomic numbers and atomic mass numbers of the original elements in P and Q.