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Chapter 31: Q24PE (page 1149)

α decay of226Ra ,another isotope in the decay series of238U , 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 .

Short Answer

Expert verified

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

Step by step solution

01

What is atomic mass number?

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

02

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

03

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 \(_{88}^{226}R{a_{138}} \to _{86}^{222}R{n_{136}} + _2^4H{e_2}\).

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