The equation for radioactive decay of H 1 3 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.

Question

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Answer 1

Question

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

Answer to Problem 4E

The radioactive decay of $H13$ that produces β is depicted as $H13→H23e+e−10$ , as β is $e−10$

Explanation of Solution

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

$H13→?+e−10$

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 $H13→H23e+e−10$

(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

Answer to Problem 4E

The radioactive decay of $L38i$ which produces β and then later on a. s depicted as $L38i→2H24e+e−10$ , as β is $e−10$

Explanation of Solution

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

$L38i→?+e−10$

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 $L38i→B48e+e−10$ ............…. (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

$B48e→H24e+H24e$ ............. (2)

Adding equation (1) and (2) we get

$L38i→B48e+e−10B48e→H24e+H24eL38i→2H24e+e−10$

(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 $L38i→2H24e+e−10$

(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

Answer to Problem 4E

The radioactive decay of $B47e$ where there is electron capture has equation as $B47e+e−10→L37i$

Explanation of Solution

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

$B47e+e−10→?$

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 $B47e+e−10→L37i$

Hence, the overall equation for the radioactive decay of $B47e$ where there is electron capture is depicted as $B47e+e−10→L37i$

(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

Answer to Problem 4E

The radioactive decay of $B58$ which emits positron has equation as $B58→B48e+e10$

Explanation of Solution

The radioactive decay of $B58$ which emits positron is depicted as

$B58→x+e10$

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 $B58→B48e+e10$

Hence, the overall equation for the radioactive decay $B58$ which emits positron has equation as $B58→B48e+e10$

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Answer 2

Question

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

Answer to Problem 4E

The radioactive decay of $H13$ that produces β is depicted as $H13→H23e+e−10$ , as β is $e−10$

Explanation of Solution

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

$H13→?+e−10$

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 $H13→H23e+e−10$

(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

Answer to Problem 4E

The radioactive decay of $L38i$ which produces β and then later on a. s depicted as $L38i→2H24e+e−10$ , as β is $e−10$

Explanation of Solution

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

$L38i→?+e−10$

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 $L38i→B48e+e−10$ ............…. (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

$B48e→H24e+H24e$ ............. (2)

Adding equation (1) and (2) we get

$L38i→B48e+e−10B48e→H24e+H24eL38i→2H24e+e−10$

(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 $L38i→2H24e+e−10$

(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

Answer to Problem 4E

The radioactive decay of $B47e$ where there is electron capture has equation as $B47e+e−10→L37i$

Explanation of Solution

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

$B47e+e−10→?$

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 $B47e+e−10→L37i$

Hence, the overall equation for the radioactive decay of $B47e$ where there is electron capture is depicted as $B47e+e−10→L37i$

(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

Answer to Problem 4E

The radioactive decay of $B58$ which emits positron has equation as $B58→B48e+e10$

Explanation of Solution

The radioactive decay of $B58$ which emits positron is depicted as

$B58→x+e10$

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 $B58→B48e+e10$

Hence, the overall equation for the radioactive decay $B58$ which emits positron has equation as $B58→B48e+e10$

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Answer 3

Question

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

Answer to Problem 4E

The radioactive decay of $H13$ that produces β is depicted as $H13→H23e+e−10$ , as β is $e−10$

Explanation of Solution

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

$H13→?+e−10$

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 $H13→H23e+e−10$

(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

Answer to Problem 4E

The radioactive decay of $L38i$ which produces β and then later on a. s depicted as $L38i→2H24e+e−10$ , as β is $e−10$

Explanation of Solution

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

$L38i→?+e−10$

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 $L38i→B48e+e−10$ ............…. (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

$B48e→H24e+H24e$ ............. (2)

Adding equation (1) and (2) we get

$L38i→B48e+e−10B48e→H24e+H24eL38i→2H24e+e−10$

(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 $L38i→2H24e+e−10$

(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

Answer to Problem 4E

The radioactive decay of $B47e$ where there is electron capture has equation as $B47e+e−10→L37i$

Explanation of Solution

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

$B47e+e−10→?$

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 $B47e+e−10→L37i$

Hence, the overall equation for the radioactive decay of $B47e$ where there is electron capture is depicted as $B47e+e−10→L37i$

(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

Answer to Problem 4E

The radioactive decay of $B58$ which emits positron has equation as $B58→B48e+e10$

Explanation of Solution

The radioactive decay of $B58$ which emits positron is depicted as

$B58→x+e10$

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 $B58→B48e+e10$

Hence, the overall equation for the radioactive decay $B58$ which emits positron has equation as $B58→B48e+e10$

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Answer 4

Question

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

Answer to Problem 4E

The radioactive decay of $H13$ that produces β is depicted as $H13→H23e+e−10$ , as β is $e−10$

Explanation of Solution

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

$H13→?+e−10$

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 $H13→H23e+e−10$

(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

Answer to Problem 4E

The radioactive decay of $L38i$ which produces β and then later on a. s depicted as $L38i→2H24e+e−10$ , as β is $e−10$

Explanation of Solution

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

$L38i→?+e−10$

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 $L38i→B48e+e−10$ ............…. (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

$B48e→H24e+H24e$ ............. (2)

Adding equation (1) and (2) we get

$L38i→B48e+e−10B48e→H24e+H24eL38i→2H24e+e−10$

(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 $L38i→2H24e+e−10$

(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

Answer to Problem 4E

The radioactive decay of $B47e$ where there is electron capture has equation as $B47e+e−10→L37i$

Explanation of Solution

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

$B47e+e−10→?$

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 $B47e+e−10→L37i$

Hence, the overall equation for the radioactive decay of $B47e$ where there is electron capture is depicted as $B47e+e−10→L37i$

(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

Answer to Problem 4E

The radioactive decay of $B58$ which emits positron has equation as $B58→B48e+e10$

Explanation of Solution

The radioactive decay of $B58$ which emits positron is depicted as

$B58→x+e10$

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 $B58→B48e+e10$

Hence, the overall equation for the radioactive decay $B58$ which emits positron has equation as $B58→B48e+e10$

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Answer 5

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

Answer to Problem 4E

The radioactive decay of $H13$ that produces β is depicted as $H13→H23e+e−10$ , as β is $e−10$

Explanation of Solution

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

$H13→?+e−10$

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 $H13→H23e+e−10$

(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

Answer to Problem 4E

The radioactive decay of $L38i$ which produces β and then later on a. s depicted as $L38i→2H24e+e−10$ , as β is $e−10$

Explanation of Solution

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

$L38i→?+e−10$

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 $L38i→B48e+e−10$ ............…. (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

$B48e→H24e+H24e$ ............. (2)

Adding equation (1) and (2) we get

$L38i→B48e+e−10B48e→H24e+H24eL38i→2H24e+e−10$

(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 $L38i→2H24e+e−10$

(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

Answer to Problem 4E

The radioactive decay of $B47e$ where there is electron capture has equation as $B47e+e−10→L37i$

Explanation of Solution

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

$B47e+e−10→?$

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 $B47e+e−10→L37i$

Hence, the overall equation for the radioactive decay of $B47e$ where there is electron capture is depicted as $B47e+e−10→L37i$

(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

Answer to Problem 4E

The radioactive decay of $B58$ which emits positron has equation as $B58→B48e+e10$

Explanation of Solution

The radioactive decay of $B58$ which emits positron is depicted as

$B58→x+e10$

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 $B58→B48e+e10$

Hence, the overall equation for the radioactive decay $B58$ which emits positron has equation as $B58→B48e+e10$

Answer 6

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

Answer to Problem 4E

The radioactive decay of $H13$ that produces β is depicted as $H13→H23e+e−10$ , as β is $e−10$

Explanation of Solution

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

$H13→?+e−10$

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 $H13→H23e+e−10$

Answer 7

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.

Answer 8

(a)

Expert Solution

Answer to Problem 4E

The radioactive decay of $H13$ that produces β is depicted as $H13→H23e+e−10$ , as β is $e−10$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

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

$H13→?+e−10$

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 $H13→H23e+e−10$

Answer 13

(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

Answer to Problem 4E

The radioactive decay of $L38i$ which produces β and then later on a. s depicted as $L38i→2H24e+e−10$ , as β is $e−10$

Explanation of Solution

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

$L38i→?+e−10$

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 $L38i→B48e+e−10$ ............…. (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

$B48e→H24e+H24e$ ............. (2)

Adding equation (1) and (2) we get

$L38i→B48e+e−10B48e→H24e+H24eL38i→2H24e+e−10$

(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 $L38i→2H24e+e−10$

Answer 14

(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.

Answer 15

(b)

Expert Solution

Answer to Problem 4E

The radioactive decay of $L38i$ which produces β and then later on a. s depicted as $L38i→2H24e+e−10$ , as β is $e−10$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

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

$L38i→?+e−10$

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 $L38i→B48e+e−10$ ............…. (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

$B48e→H24e+H24e$ ............. (2)

Adding equation (1) and (2) we get

$L38i→B48e+e−10B48e→H24e+H24eL38i→2H24e+e−10$

(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 $L38i→2H24e+e−10$

Answer 20

(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

Answer to Problem 4E

The radioactive decay of $B47e$ where there is electron capture has equation as $B47e+e−10→L37i$

Explanation of Solution

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

$B47e+e−10→?$

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 $B47e+e−10→L37i$

Hence, the overall equation for the radioactive decay of $B47e$ where there is electron capture is depicted as $B47e+e−10→L37i$

Answer 21

(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.

Answer 22

(c)

Expert Solution

Answer to Problem 4E

The radioactive decay of $B47e$ where there is electron capture has equation as $B47e+e−10→L37i$

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

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

$B47e+e−10→?$

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 $B47e+e−10→L37i$

Hence, the overall equation for the radioactive decay of $B47e$ where there is electron capture is depicted as $B47e+e−10→L37i$

Answer 27

(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

Answer to Problem 4E

The radioactive decay of $B58$ which emits positron has equation as $B58→B48e+e10$

Explanation of Solution

The radioactive decay of $B58$ which emits positron is depicted as

$B58→x+e10$

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 $B58→B48e+e10$

Hence, the overall equation for the radioactive decay $B58$ which emits positron has equation as $B58→B48e+e10$

Answer 28

(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.

Answer 29

(d)

Expert Solution

Answer to Problem 4E

The radioactive decay of $B58$ which emits positron has equation as $B58→B48e+e10$

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

The radioactive decay of $B58$ which emits positron is depicted as

$B58→x+e10$

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 $B58→B48e+e10$