Question

Consider the following nuclear equation involving the element Mercury.200/80Hg + =201/80HgA. Identify the particle that reacted with Mercury 200/80B Why did the symbol of Mercury not change in this reaction?

Answer 1

A. Neutron

B. There is no change in the number of protons

Procedure

Consider the following chemical equation

`_(80)^(200)\text{Hg }+\text{ }=_(^80)^(201)\text{Hg}`

A. The reaction can happen if there is a neutron that collides with the isotope with a weight of 200 amu, this is because the charge remains constant, therefore it will not be feasible to preserve a neutral charge in case a proton or an electron collides.

In the equation, this is written

`_(80)^(200)\text{Hg }+_0^1\text{ n }=_(^80)^(201)\text{Hg}`

B. Because the protons remain the same in the nucleus, the number of protons determines the nature of the element, mercury will always have 80 protons in its nucleus, if this changes we will be speaking of another element.