Nuclear reaction – What is the asterisk an indication of?
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I've been observing the two-step decay of iodine-131 and can't quite determine what the asterisk above Xenon-131 indicates.
I'm currently unsure as to what precisely the superscript asterisk on Xenon is. Apologies for the meekly idiotic inquiry, I'm not all too educated on this matter.
electrons radiation radioactivity gamma-rays half-life
edited Nov 25 at 23:03
AccidentalFourierTransform
24.9k1467122
asked Nov 25 at 17:50
Skander Lejmi
182
add a comment |
up vote
3
down vote
favorite
I've been observing the two-step decay of iodine-131 and can't quite determine what the asterisk above Xenon-131 indicates.
I'm currently unsure as to what precisely the superscript asterisk on Xenon is. Apologies for the meekly idiotic inquiry, I'm not all too educated on this matter.
electrons radiation radioactivity gamma-rays half-life
edited Nov 25 at 23:03
AccidentalFourierTransform
24.9k1467122
asked Nov 25 at 17:50
Skander Lejmi
182
On a related note: en.wikipedia.org/wiki/Nuclear_isomer
– PM 2Ring
Nov 25 at 18:23
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I've been observing the two-step decay of iodine-131 and can't quite determine what the asterisk above Xenon-131 indicates.
I'm currently unsure as to what precisely the superscript asterisk on Xenon is. Apologies for the meekly idiotic inquiry, I'm not all too educated on this matter.
electrons radiation radioactivity gamma-rays half-life
edited Nov 25 at 23:03
AccidentalFourierTransform
24.9k1467122
asked Nov 25 at 17:50
Skander Lejmi
182
I've been observing the two-step decay of iodine-131 and can't quite determine what the asterisk above Xenon-131 indicates.
I'm currently unsure as to what precisely the superscript asterisk on Xenon is. Apologies for the meekly idiotic inquiry, I'm not all too educated on this matter.
electrons radiation radioactivity gamma-rays half-life
electrons radiation radioactivity gamma-rays half-life
edited Nov 25 at 23:03
AccidentalFourierTransform
24.9k1467122
asked Nov 25 at 17:50
Skander Lejmi
182
edited Nov 25 at 23:03
AccidentalFourierTransform
24.9k1467122
asked Nov 25 at 17:50
Skander Lejmi
182
edited Nov 25 at 23:03
AccidentalFourierTransform
24.9k1467122
edited Nov 25 at 23:03
AccidentalFourierTransform
24.9k1467122
edited Nov 25 at 23:03
AccidentalFourierTransform
24.9k1467122
24.9k1467122
asked Nov 25 at 17:50
Skander Lejmi
182
asked Nov 25 at 17:50
Skander Lejmi
182
asked Nov 25 at 17:50
Skander Lejmi
182
182
On a related note: en.wikipedia.org/wiki/Nuclear_isomer
– PM 2Ring
Nov 25 at 18:23
add a comment |
On a related note: en.wikipedia.org/wiki/Nuclear_isomer
– PM 2Ring
Nov 25 at 18:23
On a related note: en.wikipedia.org/wiki/Nuclear_isomer
– PM 2Ring
Nov 25 at 18:23
On a related note: en.wikipedia.org/wiki/Nuclear_isomer
– PM 2Ring
Nov 25 at 18:23
add a comment |
1 Answer
1
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votes
up vote
5
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It means that the nucleus is not in its ground state but in an excited one. As such, Xe* can and does further decay into the ground state of Xe.
answered Nov 25 at 18:02
chiappette
405215
many thanks for the clarification
– Skander Lejmi
Nov 25 at 18:12
2
More precisely, since it is a nuclear reaction, the nucleus is not in the ground state.
– Jon Custer
Nov 25 at 18:22
You're much right, I'll correct my answer
– chiappette
Nov 25 at 21:42
It may be worth noting that occasionally this means the first excited state in particular (with $mathrm{Xe}^{**}$ meaning the second state and $mathrm{Xe}^{***}$ the third, etc) but that that the same notation is often used in an inclusive manner to mean any excited state. Which meaning is intended must generally be extracted from the context.
– dmckee♦
Nov 25 at 22:48
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
It means that the nucleus is not in its ground state but in an excited one. As such, Xe* can and does further decay into the ground state of Xe.
answered Nov 25 at 18:02
chiappette
405215
many thanks for the clarification
– Skander Lejmi
Nov 25 at 18:12
2
More precisely, since it is a nuclear reaction, the nucleus is not in the ground state.
– Jon Custer
Nov 25 at 18:22
You're much right, I'll correct my answer
– chiappette
Nov 25 at 21:42
It may be worth noting that occasionally this means the first excited state in particular (with $mathrm{Xe}^{**}$ meaning the second state and $mathrm{Xe}^{***}$ the third, etc) but that that the same notation is often used in an inclusive manner to mean any excited state. Which meaning is intended must generally be extracted from the context.
– dmckee♦
Nov 25 at 22:48
add a comment |
up vote
5
down vote
accepted
It means that the nucleus is not in its ground state but in an excited one. As such, Xe* can and does further decay into the ground state of Xe.
answered Nov 25 at 18:02
chiappette
405215
many thanks for the clarification
– Skander Lejmi
Nov 25 at 18:12
2
More precisely, since it is a nuclear reaction, the nucleus is not in the ground state.
– Jon Custer
Nov 25 at 18:22
You're much right, I'll correct my answer
– chiappette
Nov 25 at 21:42
It may be worth noting that occasionally this means the first excited state in particular (with $mathrm{Xe}^{**}$ meaning the second state and $mathrm{Xe}^{***}$ the third, etc) but that that the same notation is often used in an inclusive manner to mean any excited state. Which meaning is intended must generally be extracted from the context.
– dmckee♦
Nov 25 at 22:48
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
It means that the nucleus is not in its ground state but in an excited one. As such, Xe* can and does further decay into the ground state of Xe.
answered Nov 25 at 18:02
chiappette
405215
It means that the nucleus is not in its ground state but in an excited one. As such, Xe* can and does further decay into the ground state of Xe.
answered Nov 25 at 18:02
chiappette
405215
edited Nov 25 at 21:43
answered Nov 25 at 18:02
chiappette
405215
answered Nov 25 at 18:02
chiappette
405215
answered Nov 25 at 18:02
chiappette
405215
405215
many thanks for the clarification
– Skander Lejmi
Nov 25 at 18:12
2
More precisely, since it is a nuclear reaction, the nucleus is not in the ground state.
– Jon Custer
Nov 25 at 18:22
You're much right, I'll correct my answer
– chiappette
Nov 25 at 21:42
It may be worth noting that occasionally this means the first excited state in particular (with $mathrm{Xe}^{**}$ meaning the second state and $mathrm{Xe}^{***}$ the third, etc) but that that the same notation is often used in an inclusive manner to mean any excited state. Which meaning is intended must generally be extracted from the context.
– dmckee♦
Nov 25 at 22:48
add a comment |
many thanks for the clarification
– Skander Lejmi
Nov 25 at 18:12
2
More precisely, since it is a nuclear reaction, the nucleus is not in the ground state.
– Jon Custer
Nov 25 at 18:22
You're much right, I'll correct my answer
– chiappette
Nov 25 at 21:42
It may be worth noting that occasionally this means the first excited state in particular (with $mathrm{Xe}^{**}$ meaning the second state and $mathrm{Xe}^{***}$ the third, etc) but that that the same notation is often used in an inclusive manner to mean any excited state. Which meaning is intended must generally be extracted from the context.
– dmckee♦
Nov 25 at 22:48
many thanks for the clarification
– Skander Lejmi
Nov 25 at 18:12
many thanks for the clarification
– Skander Lejmi
Nov 25 at 18:12
2
2
More precisely, since it is a nuclear reaction, the nucleus is not in the ground state.
– Jon Custer
Nov 25 at 18:22
More precisely, since it is a nuclear reaction, the nucleus is not in the ground state.
– Jon Custer
Nov 25 at 18:22
You're much right, I'll correct my answer
– chiappette
Nov 25 at 21:42
You're much right, I'll correct my answer
– chiappette
Nov 25 at 21:42
It may be worth noting that occasionally this means the first excited state in particular (with $mathrm{Xe}^{**}$ meaning the second state and $mathrm{Xe}^{***}$ the third, etc) but that that the same notation is often used in an inclusive manner to mean any excited state. Which meaning is intended must generally be extracted from the context.
– dmckee♦
Nov 25 at 22:48
It may be worth noting that occasionally this means the first excited state in particular (with $mathrm{Xe}^{**}$ meaning the second state and $mathrm{Xe}^{***}$ the third, etc) but that that the same notation is often used in an inclusive manner to mean any excited state. Which meaning is intended must generally be extracted from the context.
– dmckee♦
Nov 25 at 22:48
add a comment |
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On a related note: en.wikipedia.org/wiki/Nuclear_isomer
– PM 2Ring
Nov 25 at 18:23