Q24PE - College Physics (Urone)

Question

α decay of ²²⁶Ra ,another isotope in the decay series of ²³⁸U , first recognized as a new element by the Curies. Poses special problems because its daughter is a radioactive noble gas. In the following four problems, identify the parent nuclide and write the complete decay equation in the \(_{\rm{Z}}^{\rm{A}}{{\rm{X}}_{\rm{N}}}\) notation. Refer to the periodic table for values of Z .

Step-by-Step Solution

Answer

The alpha Decay equation of ²²⁶Ra is \(_{88}^{226}R{a_{138}} \to _{86}^{222}R{n_{136}} + _2^4H{e_2}\).

What is atomic mass number?

The amount of matter contained in an atom of an element is called its atomic mass.

Formula to be used

A = N + Z

 

Where A is atomic mass number

Z is the number of protons in a nucleus

X is the symbol for the element

 

In the expression below:

\(_Z^A{X_N}\) 

 

Z is the number of protons in a nucleus

X is the symbol for the element

To determine the alpha decay equation of

²²⁶Ra

We know that 

A = N + Z  

 

Where A is atomic mass number

 

The atomic mass of \(_{88}^{226}R{a_{138}}\) is 226 and

A = 226 Z = 88 N = 138

 

Thus,

 

A = 222 N = 138 - 2 = 136 Z = 88 - 2 = 86

 

Therefore, alpha Decay equation of ²²⁶Ra is \(_{88}^{226}R{a_{138}} \to _{86}^{222}R{n_{136}} + _2^4H{e_2}\).