Chapter 19, Problem 22QAP

(a)

Interpretation Introduction

Interpretation: The given nuclear equation should be completed.

  ${{}^{196}}_{85}At{\to }^4{}_{2}He +?$

Concept Introduction: When an atom on undergoing radioactive decay emits an alpha particle (helium nucleus, ⁴₂He) from the nucleus of an atom, then such type of radioactive decays is said to be alpha decay.

Answer to Problem 22QAP

  ${{}^{196}}_{85}At{\to }^4{}_{2}He {+}^{192}{}_{83}Bi$

Explanation of Solution

Given:

  ${{}^{196}}_{85}At{\to }^4{}_{2}He +?$

When an atom on undergoing radioactive decay emits an alpha particle (helium nucleus, ⁴₂He) from the nucleus of an atom, then such type of radioactive decays is said to be alpha decay.

The general reaction for alpha decay is:

  ${{}^A}_{Z}X {\to }^{A-4}{}_{Z-2}Y{+}^4{}_{2}He$

Where A is mass number, Z is atomic number (number of protons), and X is the symbol of element.

The given atom is At-196.

The atomic number of At is 85. The alpha emission will result in decrease in number of protons by 2 so, the number of protons of the element formed from beta emission of At-196will be 83 and the mass number will decrease by 4 so, the mass number of elements is 196 − 4 = 192. The element with atomic number 83 is Bismuth, Bi. Thus, the complete nuclear reaction is:

  ${{}^{196}}_{85}At{\to }^4{}_{2}He {+}^{192}{}_{83}Bi$

(b)

Interpretation Introduction

Interpretation: The given nuclear equation should be completed.

  ${{}^{208}}_{84}Po{\to }^4{}_{2}He+?$

Concept Introduction: When an atom on undergoing radioactive decay emits an alpha particle (helium nucleus, ⁴₂He) from the nucleus of an atom, then such type of radioactive decays is said to be alpha decay.

Answer to Problem 22QAP

  ${{}^{208}}_{84}Po{\to }^4{}_{2}He{+}^{204}{}_{82}Pb$

Explanation of Solution

Given:

  ${{}^{208}}_{84}Po{\to }^4{}_{2}He+?$

When an atom on undergoing radioactive decay emits an alpha particle (helium nucleus, ⁴₂He) from the nucleus of an atom then such type of radioactive decays is said to be alpha decay.

The general reaction for alpha decay is:

  ${{}^A}_{Z}X {\to }^{A-4}{}_{Z-2}Y{+}^4{}_{2}He$

Where A is mass number, Z is atomic number (number of protons), and X is the symbol of element.

The given atom is Po-208.

The atomic number of Po is 84. The alpha emission will result in decrease in number of protons by 2 so, the number of protons of the element formed from beta emission of Po-208 will be 82 and the mass number will decrease by 4 so, the mass number of elements is 208 − 4 = 204. The element with atomic number 82 is Lead, Pb. Thus, the complete nuclear reaction is:

  ${{}^{208}}_{84}Po{\to }^4{}_{2}He{+}^{204}{}_{82}Pb$

(c)

Interpretation Introduction

Interpretation: The given nuclear equation should be completed.

  ${{}^{210}}_{86}Rn{\to }^4{}_{2}He+?$

Concept Introduction: When an atom on undergoing radioactive decay emits an alpha particle (helium nucleus, ⁴₂He) from the nucleus of an atom then such type of radioactive decays is said to be alpha decay.

Answer to Problem 22QAP

  ${{}^{210}}_{86}Rn{\to }^4{}_{2}He{+}^{206}{}_{84}Po$

Explanation of Solution

Given:

  ${{}^{210}}_{86}Rn{\to }^4{}_{2}He+?$

When an atom on undergoing radioactive decay emits an alpha particle (helium nucleus, ⁴₂He) from the nucleus of an atom, then such type of radioactive decays is said to be alpha decay.

The general reaction for alpha decay is:

  ${{}^A}_{Z}X {\to }^{A-4}{}_{Z-2}Y{+}^4{}_{2}He$

Where A is mass number, Z is atomic number (number of protons), and X is the symbol of element.

The given atom is Rn-210.

The atomic number of Rn is 86. The alpha emission will result in decrease in number of protons by 2 so, the number of protons of the element formed from beta emission of Rn-210 will be 84 and the mass number will decrease by 4 so, the mass number of elements is 210 − 4 = 206. The element with atomic number 84 is Polonium, Po. Thus, the complete nuclear reaction is:

  ${{}^{210}}_{86}Rn{\to }^4{}_{2}He{+}^{206}{}_{84}Po$