Chapter 25, Problem 1E

Interpretation Introduction

(a)

Interpretation:

The given radioactive process needs to be explained.

${\beta }^{-}$ emission of ${}_{26}^{55}\text{Fe}$ isotope.

Concept introduction:

The emission of electromagnetic radiation or particles from nucleus takes place in radioactive processes. The common radiations emitted by nucleus are as follows:

(a)Alpha decay ${}_2^4\text{He}$

(b) Beta decay ${}_{-1}^0\beta$

(c) Gamma decay ${}_0^0\gamma$

Answer to Problem 1E

${}_{26}^{55}\text{Fe}{\to }_{25}^{55}{\text{Fe+}}_{-1}^0\beta$

Explanation of Solution

In ${\beta }^{-}$ emission, an electron is emitted from the nucleus. The given nucleus is ${}_{26}^{55}\text{Fe}$ here, atomic number is 26 and mass number is 55.

Atomic number is equal to number of protons or number of electrons. Also, mass number is sum of number of protons and neutrons.

In beta decay, the number of electrons decreases thus, atomic number decreases to 25 but mass number remains the same.

${}_{26}^{55}\text{Fe}{\to }_{25}^{55}{\text{Fe+}}_{-1}^0\beta$

Interpretation Introduction

(b)

Interpretation:

The given radioactive process needs to be explained.

Alpha decay of ${}_{92}^{238}\text{U}$

Concept introduction:

The emission of electromagnetic radiation or particles from nucleus takes place in radioactive processes. The common radiations from emitted by nucleus are as follows:

(a)Alpha decay ${}_2^4\text{He}$

(b)Beta decay ${}_{-1}^0\beta$

(c)Gamma decay ${}_0^0\gamma$

Answer to Problem 1E

${}_{92}^{238}\text{U}{\to }_{90}^{234}{\text{Th+}}_2^{4}He$

Explanation of Solution

In $\alpha$ decay, emission of helium atom ${}_2^4\text{He}$ takes place. The given nucleus is ${}_{92}^{238}\text{U}$ here, atomic number is 92 and mass number is 238.

Atomic number is equal to number of protons or number of electrons. Also, mass number is sum of number of protons and neutrons.

In alpha decay, the atomic number gets decreases by 2 unit and mass number decreases by 4 units thus, the new nucleus formed will have 234 as mass number and 90 as atomic number.

The element corresponding to atomic number 90 is thorium.

The nuclear reaction can be represented as follows:

${}_{92}^{238}\text{U}{\to }_{90}^{234}{\text{Th+}}_2^{4}He$

Interpretation Introduction

(c)

Interpretation:

The given radioactive process needs to be explained.

Two successive alpha decay of ${}_{86}^{222}\text{Rn}\text{.}$

Concept introduction:

The emission of electromagnetic radiation or particles from nucleus takes place in radioactive processes. The common radiations emitted by nucleus are as follows:

(a) Alpha decay ${}_2^4\text{He}$

(b) Beta decay ${}_{-1}^0\beta$

(c)Gamma decay ${}_0^0\gamma$

Answer to Problem 1E

${}_{86}^{222}\text{Rn}{\stackrel{{-}_2^{4}He}{\to }}_{84}^{218}\text{Po}{\stackrel{{-}_2^{4}He}{\to }}_{82}^{214}\text{Pb}$

Explanation of Solution

In $\alpha$ decay, emission of helium atom ${}_2^4\text{He}$ takes place. The given nucleus is ${}_{86}^{222}\text{Rn}$ here, atomic number is 86 and mass number is 222.

Atomic number is equal to number of protons or number of electrons. Also, mass number is the total sum of both protons and neutrons.

In alpha decay, the atomic number gets decreases by 2 unit and mass number decreases by 4 units. In first alpha decay, the new nucleus formed will have 218 as mass number and 84 as atomic number.

The element corresponding to atomic number 84 is polonium.

The nuclear reaction can be represented as follows:

${}_{86}^{222}\text{Rn}{\to }_{84}^{218}{\text{Po+}}_2^{4}He$

After second alpha decay, the new nucleus formed will have 214 as mass number and 82 as atomic number.

The element corresponding to atomic number 82 is lead.

The nuclear reaction can be represented as follows:

${}_{84}^{218}\text{Po}{\to }_{82}^{214}{\text{Pb+}}_2^{4}He$

Thus, the nuclear reaction representing two successive alpha decays is as follows:

${}_{86}^{222}\text{Rn}{\stackrel{{-}_2^{4}He}{\to }}_{84}^{218}\text{Po}{\stackrel{{-}_2^{4}He}{\to }}_{82}^{214}\text{Pb}$

Interpretation Introduction

(d)

Interpretation:

The given radioactive process needs to be explained.

Two successive ${}_{-1}^0\beta$ emission of

${}_{29}^{64}\text{Cu}$

Concept introduction:

The emission of electromagnetic radiation or particles from nucleus takes place in radioactive processes. The common radiations from emitted by nucleus are as follows:

(a) Alpha decay ${}_2^4\text{He}$

(b) Beta decay ${}_{-1}^0\beta$

(c) Gamma decay ${}_0^0\gamma$

Answer to Problem 1E

${}_{29}^{64}\text{Cu}{\stackrel{{-}_{-1}^0\beta }{\to }}_{28}^{64}\text{Ni}{\stackrel{{-}_{-1}^0\beta }{\to }}_{27}^{64}\text{Co}$

Explanation of Solution

In ${\beta }^{-}$ emission, an electron is emitted from the nucleus. The given nucleus is ${}_{29}^{64}\text{Cu}$ here, atomic number is 29 and mass number is 64.

Atomic number is equal to number of protons or number of electrons. Also, mass number is sum of number of protons and neutrons.

In beta decay, the number of electrons decreases and mass number remains the same.

In first beta decay, the new nucleus formed will have 64 as mass number and 28 as atomic number.

The element corresponding to atomic number 28 is nickel.

The nuclear reaction can be represented as follows:

${}_{29}^{64}\text{Cu}{\stackrel{}{\to }}_{28}^{64}{\text{Ni+}}_{-1}^0\beta$

After second beta decay, the new nucleus formed will have 64 as mass number and 27 as atomic number.

The element corresponding to atomic number 27 is cobalt.

The nuclear reaction can be represented as follows:

${}_{28}^{64}\text{Ni}{\stackrel{}{\to }}_{27}^{64}{\text{Co+}}_{-1}^0\beta$

Thus, the nuclear reaction representing two successive beta decays is as follows:

${}_{29}^{64}\text{Cu}{\stackrel{{-}_{-1}^0\beta }{\to }}_{28}^{64}\text{Ni}{\stackrel{{-}_{-1}^0\beta }{\to }}_{27}^{64}\text{Co}$