Chemistry: The Science in Context (Fifth Edition)
Chemistry: The Science in Context (Fifth Edition)
5th Edition
ISBN: 9780393614046
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster, Stacey Lowery Bretz, Geoffrey Davies
Publisher: W. W. Norton & Company
Question
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Chapter 19, Problem 19.10VP

(a)

Interpretation Introduction

Interpretation: The based on the given images are to be answered.

Concept introduction: The beta particle is a negatively charged particle with zero mass charge.

Fusion reaction is the one in which the number of reactants are more than the number of products.

The positron particle is a positively charged particle with zero mass charge.

Fission reaction is the one in which the number of reactants are less than the number of products.

Radon222 and americium241 undergo alpha decay.

Lead210 and iodine131 undergo beta decay.

To determine: The image that represents beta decay and a balanced nuclear equation to represent the image.

(a)

Expert Solution
Check Mark

Answer to Problem 19.10VP

Solution:

Image B represents beta decay as,

ZAXZ+1AY+10β

Explanation of Solution

The beta particle is a negatively charged particle with zero mass charge.

In beta decay, a radioactive substance decomposes to emit a beta particle along with a nuclide with the atomic number more than that of the radioactive nuclide.

From the given images, image B represents a nuclear reaction in which one substance dissociate into another substance with the removal of a negatively charge particle.

Chemistry: The Science in Context (Fifth Edition), Chapter 19, Problem 19.10VP , additional homework tip  1

Figure 1

Therefore, image B represents beta decay as,

ZAXZ+1AY+10β

(b)

Interpretation Introduction

To determine: The image that represents fusion and a balanced nuclear equation to represent the image.

(b)

Expert Solution
Check Mark

Answer to Problem 19.10VP

Solution:

Image D represents fusion as,

12H+12H+12H+12H24He+10β+10β+Energy

Explanation of Solution

Fusion reaction is the one in which the number of reactants are more than the number of products.

From the given images, image D represents a nuclear reaction in which the number of reactants is four and the number of products is three with the release of energy.

If represents the fusion of four hydrogen atoms to form a helium atom and two beta particles.

Chemistry: The Science in Context (Fifth Edition), Chapter 19, Problem 19.10VP , additional homework tip  2

Figure 2

Therefore, image D represents fusion as,

12H+12H+12H+12H24He+10β+10β+Energy

(c)

Interpretation Introduction

To determine: The image that represents positron emission and a balanced nuclear equation to represent the image.

(c)

Expert Solution
Check Mark

Answer to Problem 19.10VP

Solution:

Image B represents positron emission as,

ZAXZ+1AY+10β

Explanation of Solution

The positron particle is a positively charged particle with zero mass charge.

In positron emission, a radioactive substance decomposes to emit a positron particle along with a nuclide with the atomic number less than that of the radioactive nuclide.

From the given images, image H represents a nuclear reaction in which one substance dissociate into another substance with the removal of a positively charge particle.

Chemistry: The Science in Context (Fifth Edition), Chapter 19, Problem 19.10VP , additional homework tip  3

Figure 3

Therefore, image H represents positron emission as,

ZAXZ+1AY+10β

(d)

Interpretation Introduction

To determine: The image that represents fission and a balanced nuclear equation to represent the image.

(d)

Expert Solution
Check Mark

Answer to Problem 19.10VP

Solution:

Image F represents fission as,

92233U+01n54137Xe+3894Sr+301n

Explanation of Solution

Fission reaction is the one in which the number of reactants are less than the number of products.

From the given images, image F represents a nuclear reaction in which two reactants break into six particles.

If represents the fission of uranium233 reacting with neutron to form a xenon isotope, strontium isotope and three neutrons.

Chemistry: The Science in Context (Fifth Edition), Chapter 19, Problem 19.10VP , additional homework tip  4

Figure 4

Therefore, image F represents fission as,

92233U+01n54137Xe+3894Sr+301n

(e)

Interpretation Introduction

To determine: The image that represents a nuclide that undergoes alpha decay and a balanced nuclear equation to represent the alpha decay of the nuclide.

(e)

Expert Solution
Check Mark

Answer to Problem 19.10VP

Solution:

Image A and image I represents the nuclides that undergo alpha decay.

Radon222 undergoes alpha decay as,

86222Rn84218Po+24α

Americium241 undergoes alpha decay as,

95241Am93237Np+24α+γ

Explanation of Solution

Out of the given nuclides, the one that undergoes alpha decay are radon222 and americium241 .

Image A represents radon222 and image I represents americium241 .

Chemistry: The Science in Context (Fifth Edition), Chapter 19, Problem 19.10VP , additional homework tip  5

Figure 5

Chemistry: The Science in Context (Fifth Edition), Chapter 19, Problem 19.10VP , additional homework tip  6

Figure 6

Thus, image A and image I represents the nuclides that undergo alpha decay.

In alpha decay, the mass number of the radionuclide reduces by four and the atomic number reduces by two.

Therefore, radon222 undergoes alpha decay as,

86222Rn84218Po+24α

Americium241 undergoes alpha decay as,

95241Am93237Np+24α+γ

(f)

Interpretation Introduction

To determine: The image that represents a nuclide that undergoes beta decay and a balanced nuclear equation to represent the beta decay of the nuclide.

(f)

Expert Solution
Check Mark

Answer to Problem 19.10VP

Solution:

Image C and image G represents the nuclides that undergo beta decay.

Lead210 and undergoes beta decay as,

82210Pb83210Bi+10β

Iodine131 undergoes beta decay as,

53131I54131Xe+10β

Explanation of Solution

Out of the given nuclides, the one that undergoes beta decay are lead210 and iodine131 .

Image C represents lead210 and image G represents iodine131 .

Chemistry: The Science in Context (Fifth Edition), Chapter 19, Problem 19.10VP , additional homework tip  7

Figure 7

Chemistry: The Science in Context (Fifth Edition), Chapter 19, Problem 19.10VP , additional homework tip  8

Figure 8

Thus, image C and image G represents the nuclides that undergo beta decay.

In beta decay, the mass number of the radionuclide remains same and the atomic number increases by one.

Therefore, lead210 undergoes beta decay as,

82210Pb83210Bi+10β

Iodine131 undergoes beta decay as,

53131I54131Xe+10β

Conclusion:

  1. a. Image B represents beta decay as,

    ZAXZ+1AY+10β

  2. b. Image D represents fusion as,

    12H+12H+12H+12H24He+10β+10β+Energy

  3. c. Image B represents positron emission as,

    ZAXZ+1AY+10β

  4. d. Image F represents fission as,

    92233U+01n54137Xe+3894Sr+301n

  5. e. Image A and image I represents the nuclides that undergo alpha decay.

    Radon222 undergoes alpha decay as,

    86222Rn84218Po+24α

    Americium241 undergoes alpha decay as,

    95241Am93237Np+24α+γ

  6. f. Image C and image G represents the nuclides that undergo beta decay.

    Lead210 and undergoes beta decay as,

    82210Pb83210Bi+10β

    Iodine131 undergoes beta decay as,

    53131I54131Xe+10β

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Chapter 19 Solutions

Chemistry: The Science in Context (Fifth Edition)

Ch. 19 - Prob. 19.5VPCh. 19 - Prob. 19.6VPCh. 19 - Prob. 19.7VPCh. 19 - Prob. 19.8VPCh. 19 - Prob. 19.9VPCh. 19 - Prob. 19.10VPCh. 19 - Prob. 19.11QPCh. 19 - Prob. 19.12QPCh. 19 - Prob. 19.13QPCh. 19 - Prob. 19.14QPCh. 19 - Prob. 19.15QPCh. 19 - Prob. 19.16QPCh. 19 - Prob. 19.17QPCh. 19 - Prob. 19.18QPCh. 19 - Prob. 19.19QPCh. 19 - Prob. 19.20QPCh. 19 - Prob. 19.21QPCh. 19 - Prob. 19.22QPCh. 19 - Prob. 19.23QPCh. 19 - Prob. 19.24QPCh. 19 - Prob. 19.25QPCh. 19 - Prob. 19.26QPCh. 19 - Prob. 19.27QPCh. 19 - Prob. 19.28QPCh. 19 - Prob. 19.29QPCh. 19 - Prob. 19.30QPCh. 19 - Prob. 19.31QPCh. 19 - Prob. 19.32QPCh. 19 - Prob. 19.33QPCh. 19 - Prob. 19.34QPCh. 19 - Prob. 19.35QPCh. 19 - Prob. 19.36QPCh. 19 - Prob. 19.37QPCh. 19 - Prob. 19.38QPCh. 19 - Prob. 19.39QPCh. 19 - Prob. 19.40QPCh. 19 - Prob. 19.41QPCh. 19 - Prob. 19.42QPCh. 19 - Prob. 19.43QPCh. 19 - Prob. 19.44QPCh. 19 - Prob. 19.45QPCh. 19 - Prob. 19.46QPCh. 19 - Prob. 19.47QPCh. 19 - Prob. 19.48QPCh. 19 - Prob. 19.49QPCh. 19 - Prob. 19.50QPCh. 19 - Prob. 19.51QPCh. 19 - Prob. 19.52QPCh. 19 - Prob. 19.53QPCh. 19 - Prob. 19.54QPCh. 19 - Prob. 19.55QPCh. 19 - Prob. 19.56QPCh. 19 - Prob. 19.57QPCh. 19 - Prob. 19.58QPCh. 19 - Prob. 19.59QPCh. 19 - Prob. 19.60QPCh. 19 - Prob. 19.61QPCh. 19 - Prob. 19.62QPCh. 19 - Prob. 19.63QPCh. 19 - Prob. 19.64QPCh. 19 - Prob. 19.65QPCh. 19 - Prob. 19.66QPCh. 19 - Prob. 19.67QPCh. 19 - Prob. 19.68QPCh. 19 - Prob. 19.69QPCh. 19 - Prob. 19.70QPCh. 19 - Prob. 19.71QPCh. 19 - Prob. 19.72QPCh. 19 - Prob. 19.73QPCh. 19 - Prob. 19.74QPCh. 19 - Prob. 19.75QPCh. 19 - Prob. 19.76QPCh. 19 - Prob. 19.77QPCh. 19 - Prob. 19.78QPCh. 19 - Prob. 19.79QPCh. 19 - Prob. 19.80QPCh. 19 - Prob. 19.81QPCh. 19 - Prob. 19.82QPCh. 19 - Prob. 19.83QPCh. 19 - Prob. 19.84QPCh. 19 - Prob. 19.85APCh. 19 - Prob. 19.86APCh. 19 - Prob. 19.87APCh. 19 - Prob. 19.88APCh. 19 - Prob. 19.89APCh. 19 - Prob. 19.90APCh. 19 - Prob. 19.91APCh. 19 - Prob. 19.92APCh. 19 - Prob. 19.93APCh. 19 - Prob. 19.94APCh. 19 - Prob. 19.95APCh. 19 - Prob. 19.96APCh. 19 - Prob. 19.97APCh. 19 - Prob. 19.98APCh. 19 - Prob. 19.99APCh. 19 - Prob. 19.100AP
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