Question

Calculate the number of neutrons present in the nucleus of an element $X$ which is represented as ${}_{15}^{31}X$ .

Answer 1

Hint: We have to know that the neutron is a sort of hadron. It comprises one up quark and two down quarks. Albeit the mass of a proton and a neutron are equivalent, particularly contrasted and the a lot lighter electron, a neutron is marginally more monstrous than a proton.

Complete answer:
We have to know that the proton is a subatomic molecule, image p, with a positive electric charge of ${}^{ + 1}e$ rudimentary charge and a mass somewhat not exactly that of a neutron. Protons and neutrons, each with masses of around one nuclear mass unit, are together, alluded to as "nucleons" (particles present in nuclear cores). At least one proton are available in the core of each particle; they are an essential piece of the core. The quantity of protons in the core is the characterizing property of a component, and is alluded to as the nuclear number (represented by $Z$ ). Since every component has a special number of protons, every component has its own novel nuclear number.
As per Bohr's nuclear hypothesis, an iota is made out of a core at the middle and at least one electron pivoting around the core along various energy circles. The core is essentially made out of protons and neutrons, by and large called nucleons.
$Massnumber{\text{ (A) = No}}{\text{. of protons (Z) + No}}{\text{. of neutrons}}$
In the given compound is ${}_{15}^{31}X$ ,
Where,
$No.{\text{ of neutrons = 31 - 15 = 16}}$
Therefore, the number of neutrons is $16$ .

Note:
In view of this nearby pressing, the inner construction of protons and neutrons (all things considered called nucleons) assumes a significant part in atomic physical science, it impacts, and can thusly be altered by, the conveyance, movement, and connections of nucleons inside atomic matter.