The immediate decay product of element 118 is thought to be element \(116 .\) Write a complete nuclear equation for this reaction.

Short Answer

Expert verified
\(E(118,A) \rightarrow E'(116,A-4) + He(2,4)\)

Step by step solution

01

Identify the Alpha Particle

An alpha particle is a helium nucleus, consisting of 2 protons and 2 neutrons, hence it has an atomic number of 2 and a mass number of 4.
02

Write the Decay Reaction

Using the notation \(X(Z,A)\) where X is the chemical symbol, Z is the atomic number and A is the mass number. The parent nuclide is \(E(118,A)\) and it decays into \(E'(116,A-4)\) with the emission of an alpha particle, which can be represented as \(He(2,4)\). Thus the decay reaction is: \[E(118,A) \rightarrow E'(116,A-4) + He(2,4)\]
03

Check The Balance of The Equation in Both Sides

For the nuclear equation to be correct, the sum of the mass numbers and the atomic numbers must be equal on both sides of the equation.

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