The decay of 241/95 Am by alpha particle emission can be represented by the following nuclear reaction:
241/95 Am --> 4/2 He + 237/93 Np
In this reaction, an alpha particle (4/2 He) is emitted from the nucleus of 241/95 Am, resulting in the formation of a new nucleus. The new nucleus is 237/93 Np, which has 93 protons (the same as the original Am nucleus) and 237 - 93 = 144 neutrons (four less than the original Am nucleus).
Alpha particle emission is a common mode of radioactive decay, particularly for heavy elements such as Am. During alpha decay, the nucleus emits an alpha particle, which is a helium nucleus consisting of two protons and two neutrons. This reduces the atomic number of the nucleus by two and the mass number by four.