Question

Neptunium's only naturally occurring isotope, 23793np, decays by emitting one alpha particle, one beta particle, and one gamma ray. What is the new atom formed from this decay?

Answer 1

The given isotope of Neptunium is $^{237} _{93} Np$

Alpha decay of an isotope results in daughter nuclide with mass number less by 4 units and atomic number less by 2 units than the parent isotope.

Alpha decay of Neptunium-237 can be represented as:

$^{237} _{93} Np-->^{233}_{91} Pa+^{4} _{2} He$

Beta decay of the above formed protactinium nuclide can be represented as:

$^{233}_{91}Pa-->^{233} _{92} U+^{0}_{-1}e$

Gamma decay releases only energy in the form of gamma rays, the nuclide remains the same.

$^{233}_{92}U-->^{233}_{92}U+$γ

The new atom formed after the given decays is $^{233}_{92}U$