Concept explainers
(a)
Interpretation:
The nuclide X that decays by alpha emission to give R217n is to be stated.
Concept introduction:
Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.
Answer to Problem 11E
The nuclide X that decays by alpha emission to give R217n is radium (R22188a).
Explanation of Solution
It is given that X undergoes alpha decay to give R217n. An alpha particle is H42e. Therefore, in the case of alpha decay the
Xmx→21786Rn+42He
The mass number of X can be calculated by the sum of mass number of both of the elements on the right side.
m=217+4=221
The atomic number of X can be calculated by the sum of atomic number of both of the elements on the right side.
x=86+2=88
The element having atomic number 88 is radium. Therefore, the nuclide X is represented as R22188a.
The resulting nuclear equation is shown below.
T22290h→21888Ra+42He
The nuclide X which on alpha decay gives R217n is radium (R22188a).
(b)
Interpretation:
The nuclide X that decays by beta emission to give C43a is to be stated.
Concept introduction:
Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.
Answer to Problem 11E
The nuclide X that decays by beta emission and gives C43a is K4319.
Explanation of Solution
It is given that X on beta emission gives C43a. A beta particle is identical to an electron that is e0−1. Therefore, in the case of beta decay the atomic number of an element increases by 1 and there is no change in mass number.
The nuclear equation is shown below.
Xmx→4320Ca+e0−1
The mass number of X can be calculated by the sum of mass number of both of the elements on the right side.
m=43+0=43
The atomic number of X can be calculated by the sum of atomic number of both of the elements on the right side.
x=20+(−1)=20−1=19
The element having atomic number 19 is potassium (K). Therefore, the nuclide X is represented as K4319.
The resulting nuclear equation is shown below.
K4319→4320Ca+e0−1
The nuclide X which on beta decay gives C43a is K4319.
(c)
Interpretation:
The nuclide X that decays by positron emission to give B73r is to be stated.
Concept introduction:
Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.
Answer to Problem 11E
The nuclide X that decays by positron emission and gives B73r is K7336r.
Explanation of Solution
It is given that X on positron emission gives B73r. A positron is opposite of an electron and represented by e0+1. Therefore, in the case of positron the atomic number of an element decreases by 1 and there is no change in mass number.
The nuclear equation is shown below
Xmx→7335Br+e0+1
The mass number of X can be calculated by the sum of mass number of both of the elements on the right side.
m=73+0=73
The atomic number of X can be calculated by the sum of atomic number of both of the elements on the right side.
x=35+1=36
The element having atomic number 36 is krypton (Kr). Therefore, the nuclide X is represented as K7336r.
The resulting nuclear equation is shown below.
K7336r→7335Br+e0+1
The nuclide X on positron decay gives B73r is K7336r.
(d)
Interpretation:
The nuclide X that decays by electron capture to give C133s is to be stated.
Concept introduction:
Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.
Answer to Problem 11E
The nuclide X that decays by electron capture and gives C133s is B13356a.
Explanation of Solution
It is given that X on electron capture gives C133s. Electron capture means that when an electron is used to make any unstable atom stable. Therefore, in the case of electron capture the atomic number of an element decreases by 1 and there is no change in mass number.
The nuclear equation is shown below
Xmx+e0−1→13355Cs
The mass number of X can be calculated by the sum of mass number of both of the elements on the left side.
m+0=133m=133
The atomic number of X can be calculated by the sum of atomic number of both of the elements on the left side.
x+(−1)=55x−1=55
Rearrange the above equation as shown below.
x=55+1=56
The element having atomic number 56 is barium (Ba). Therefore, the nuclide X is represented as B13356a.
The resulting nuclear equation is shown below.
B13356a+e0−1→13355Cs
The nuclide X which on electron capture gives C133s is B13356a.
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Chapter 18 Solutions
Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
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