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Chapter 21: Problem 69

Radon-222 decays to a stable nucleus by a series of three alpha emissions and two beta emissions. What is the stable nucleus that is formed?

Short Answer

Expert verified
The stable nucleus formed after three alpha emissions and two beta emissions from Radon-222 is Lead-210 (Pb-210).

Step by step solution

01

Identify the initial element and its atomic number and mass number

The initial element is Radon-222 (Rn-222). For Rn-222, - Atomic number (Z) = 86 (number of protons) - Mass number (A) = 222 (number of protons + neutrons)
02

Three alpha emissions

An alpha emission involves the loss of two protons and two neutrons from a nucleus. So, after each alpha emission, the atomic number will decrease by 2, and the mass number will decrease by 4. After the first alpha emission: - Atomic number = 86 - 2 = 84 - Mass number = 222 - 4 = 218 After the second alpha emission: - Atomic number = 84 - 2 = 82 - Mass number = 218 - 4 = 214 After the third alpha emission: - Atomic number = 82 - 2 = 80 - Mass number = 214 - 4 = 210 So at this point, we have a nucleus with an atomic number of 80 and a mass number of 210.
03

Two beta emissions

A beta emission involves the conversion of a neutron into a proton while emitting an electron (beta particle). As a result, the atomic number will increase by 1 while the mass number remains the same. After the first beta emission: - Atomic number = 80 + 1 = 81 - Mass number = 210 After the second beta emission: - Atomic number = 81 + 1 = 82 - Mass number = 210 At this point, we have a nucleus with an atomic number of 82 and a mass number of 210.
04

Identify the stable nucleus formed

The final stable nucleus has: - Atomic number (Z) = 82 - Mass number (A) = 210 This corresponds to the element Lead (Pb) with 82 protons and a mass number of 210. Therefore, the stable nucleus formed is Lead-210 (Pb-210).

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Most popular questions from this chapter

Write balanced equations for (a) \({ }_{92}^{238} \mathrm{U}(\alpha, \mathrm{n}){ }_{94}^{241} \mathrm{Pu},\) (b) \({ }_{7}^{14} \mathrm{~N}(\alpha, \mathrm{p}){ }^{17} \mathrm{O}\) (c) \({ }_{26}^{56} \mathrm{Fe}(\alpha, \beta)_{29}^{60} \mathrm{Cu}\).

Which type or types of nuclear reactors have these characteristics? (a) Does not use a secondary coolant (b) Creates more fissionable material than it consumes (c) Uses a gas, such as He or \(\mathrm{CO}_{2}\), as the primary coolant

The half-life for the process \({ }^{238} \mathrm{U} \longrightarrow{ }^{206} \mathrm{~Pb}\) is \(4.5 \times 10^{9} \mathrm{yr}\). A mineral sample contains \(75.0 \mathrm{mg}\) of \({ }^{238} \mathrm{U}\) and \(18.0 \mathrm{mg}\) of \({ }^{206} \mathrm{~Pb}\). What is the age of the mineral?

The cloth shroud from around a mummy is found to have a \({ }^{14} \mathrm{C}\) activity of 9.7 disintegrations per minute per gram of carbon as compared with living organisms that undergo 16.3 disintegrations per minute per gram of carbon. From the half-life for \({ }^{14} \mathrm{C}\) decay, \(5715 \mathrm{yr},\) calculate the age of the shroud.

Using the concept of magic numbers, explain why alpha emission is relatively common, but proton emission is nonexistent.
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