EBK CHEMICAL PRINCIPLES
EBK CHEMICAL PRINCIPLES
8th Edition
ISBN: 9781305856745
Author: DECOSTE
Publisher: CENGAGE LEARNING - CONSIGNMENT
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Chapter 20, Problem 4E

(a)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of H13 which produces β needs to be determined.

Concept Introduction:

Radioactive decay is process to convert unstable atomic nucleus to smaller stable nuclei and this is possible only when there is imbalance between the number of protons and neutrons. During this process radiations like alpha, beta, and gamma are emitted.

(a)

Expert Solution
Check Mark

Answer to Problem 4E

The radioactive decay of H13 that produces β is depicted as H13H23e+e10 , as β is e10

Explanation of Solution

The radioactive decay of H13 which produces β is depicted as below

  H13?+e10

While computing the mass numbers and atomic mass for unknown element, it is found that

Mass number: 3 = 0 +?

x = 3

Atomic number: 1 =? + (-1)

y= 1

Hence unknown element is x23 . Looking through the periodic table, it is found that element with mass number 3 is He. Hence, the element is H23e . The emission of β particle will increase the atomic number by 1 with no changes in mass number.

The radioactive decay of H13 that emits β is depicted by equation H13H23e+e10

(b)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of L38i which produces β and then later on a needs to be determined.

Concept Introduction:

When there is unstable atomic nucleus, then it gets converted to smaller stable nuclei through radioactive decay process. This happens when there is imbalance between the number of protons and neutrons. Radiations like alpha, beta, and gamma are emitted are by-products.

(b)

Expert Solution
Check Mark

Answer to Problem 4E

The radioactive decay of L38i which produces β and then later on a. s depicted as L38i2H24e+e10 , as β is e10

Explanation of Solution

The radioactive decay of L38i which produces β is depicted as below

  L38i?+e10

While computing the mass numbers and atomic mass for unknown element, it is found that

Mass number: 8 = 0 +?

x = 8

Atomic number: 3 =? + (-1)

y= 4

Hence unknown element is x48 . Looking through the periodic table, it is found that element with mass number 8 is being. Hence, the element is B48e . The emission of β particle will increase the atomic number by 1 with no changes in mass number. The radioactive decay of L38i which produces β is depicted by equation L38iB48e+e10 ............…. (1)

Then there is emission of alpha particles indicating that there will be decrease in mass number by 4 units and atomic number by 2 units. This is depicted as below

  B48eH24e+H24e ............. (2)

Adding equation (1) and (2) we get

  L38iB48e+e10B48eH24e+H24eL38i2H24e+e10

(As the B48e on both sides will cancel each other)

Hence, the overall equation for the radioactive decay of L38i which produces β and then later on a. s depicted as L38i2H24e+e10

(c)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of B47e where there is electron capture needs to be determined.

Concept Introduction:

Radioactive decay takes place when an atomic nucleus is unstable such that it gets converted to smaller stable nuclei emitting radiations like alpha, beta, and gamma. This happens only when the number of protons and neutrons are not balanced.

(c)

Expert Solution
Check Mark

Answer to Problem 4E

The radioactive decay of B47e where there is electron capture has equation as B47e+e10L37i

Explanation of Solution

The radioactive decay of B47e where there is electron capture and the equation is depicted as

  B47e+e10?

While computing the mass numbers and atomic mass for unknown element, it is found that

Mass number: 7 + 0 = ?

x = 7

Atomic number: 4 + (-1) =?

y= 3

Hence unknown element is x37 . Looking through the periodic table, it is found that element with mass number 7 and atomic number 3 is Li. Hence, the element is L37i . The electron capture will decrease the atomic number by 1 with no changes in mass number. The radioactive decay of B47e where electron capture is depicted as B47e+e10L37i

Hence, the overall equation for the radioactive decay of B47e where there is electron capture is depicted as B47e+e10L37i

(d)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of B58 which emits positron needs to be determined.

Concept Introduction:

When an atomic nucleus is unstable then it gets converted to smaller stable nuclei emitting radiations like alpha, beta, and gamma and this process is called as radioactive decay. This happens only when there is imbalance in number of protons and neutrons.

(d)

Expert Solution
Check Mark

Answer to Problem 4E

The radioactive decay of B58 which emits positron has equation as B58B48e+e10

Explanation of Solution

The radioactive decay of B58 which emits positron is depicted as

  B58x+e10

While computing the mass numbers and atomic mass for unknown element, it is found that

Mass number: 8 = x+ 0

x = 8

Atomic number: 5 = x + 1

x= 4

Hence unknown element is x48 . Looking through the periodic table, it is found that element with mass number 8 and atomic number 4 is Be. Hence, the element is B48e . The emission of positron will decrease the atomic number by 1 with no changes in mass number. The radioactive decay of B58 which emits positron is depicted as B58B48e+e10

Hence, the overall equation for the radioactive decay B58 which emits positron has equation as B58B48e+e10

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Chapter 20 Solutions

EBK CHEMICAL PRINCIPLES

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