Chapter 20, Problem 7E

(a)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of $Ga$ where electron capture takes place needs to be determined.

Concept Introduction:

Electron capture involves the absorption of electrons by the proton rich nucleus from the inner shells. During this process, the proton changes to neutron releasing an electron and atomic number of the nucleus gets reduced by 1 unit keeping the mass number the same.

Answer to Problem 7E

The radioactive decay of $Ga$ where electron capture takes place is depicted as $Ga+e\to Zn$ .

Explanation of Solution

The radioactive decay of $Ga$ where electron capture takes place is depicted as as below

  $Ga+e\to ?$

(As Ga has atomic number as 31)

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

Mass Number: 68 + 0 =?

x = 68

Atomic number: 31 + (-1) =?

y= 30

Hence unknown element is $x$ . Looking through the periodic table, it is found that element with mass number 68 and atomic number 30 is Zn. Hence, the element is $Zn$ . The electron capture releases electron in which the atomic number of the nucleus gets reduced by 1 unit keeping the mass number the same.

The radioactive decay of $Ga$ where electron capture takes place is depicted as $Ga+e\to Zn$ .

(b)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of $Cu$ which releases positron needs to be determined.

Concept Introduction:

Beta decay takes place when there is excess of protons or neutrons which gets transformed to other. In beta plus decay, proton gets converted to neutron releasing positron and neutrino. The emission of positron will decrease the atomic number by 1 with no changes in mass number.

Answer to Problem 7E

The radioactive decay of $Cu$ which produces positron is depicted as $Cu\to e+Ni$ .

Explanation of Solution

The radioactive decay of $Cu$ which produces positron is depicted as below

  $Cu\to e+?$

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

Mass number: 62 = 0 +?

x = 62

Atomic Number: 29 =? + (1)

y= 28

Hence unknown element is $x$ . Looking through the periodic table, it is found that element with mass number 62 and atomic number 28 is Ni. Hence, the element is $Ni$ . The emission of positron will decrease the atomic number by 1 with no changes in mass number. The radioactive decay of $Cu$ which produces positron is depicted as $Cu\to e+Ni$ .

Hence, the overall equation for the radioactive decay of $Cu$ which produces positron is depicted as $Cu\to e+Ni$ .

(c)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of $Fr$ where there is alpha particle release 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. Release of alpha particle causes a decrease in mass number by 4 units and atomic number by 2 units.

Answer to Problem 7E

The radioactive decay of $Fr$ with release of alpha particles has equation as $Fr\to Ar+He$ .

Explanation of Solution

The radioactive decay of $Fr$ where alpha particles are released and the equation is depicted as

  $Fr\to ?+He$

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

Mass number: 212 = ?+4

x = 208

Atomic number: 87=? + 2

y= 85

Hence unknown element is $x$ . Looking through the periodic table, it is found that element with mass number 208 and atomic number 85 is Are Hence, the element is $Ar$ . The emission of alpha particle causes a decrease in mass number by 4 units and atomic number by 2 units.

The radioactive decay of $Fr$ with release of alpha particles is depicted as $Fr\to Ar+He$ .

Hence, the overall equation for the radioactive decay of $Fr$ with release of alpha particles has equation as $Fr\to Ar+He$ .

(d)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of $Sb$ which emits beta particles needs to be explained.

Concept Introduction:

Radioactive decay releases radiations like alpha, beta and gamma, when unstable nuclei get converted to smaller stable fragments. The emission of β particle will increase the atomic number by 1 with no changes in mass number.

Answer to Problem 7E

The radioactive decay of $Sb$ which emits beta particles has equation as $Sb\to Te+e$ .

Explanation of Solution

The radioactive decay of $Sb$ which emits beta particle is depicted as

  $Sb\to ?+e$

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

Mass number: 129 = x+ 0

x = 129

Atomic number: 51 = x + (- 1)

x= 52

Hence unknown element is $x$ . Looking through the periodic table, it is found that element with mass number129 and atomic number 52 is Te. Hence, the element is $Te$ . The emission of beta particle will increase the atomic number by 1 with no changes in mass number. The radioactive decay of $Sb$ which emits beta particles has equation as $Sb\to Te+e$ .

Hence, the overall equation for the radioactive decay of $Sb$ which emits beta particles has equation as $Sb\to Te+e$ .