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Gamma decay

Gamma decay is the simplest form of nuclear decay—it is the emission of energetic photons by nuclei left in an excited state by some earlier process. Protons and neutrons in an excited nucleus are in higher orbitals, and they fall to lower levels by photon emission (analogous to electrons in excited atoms). Nuclear excited states have lifetimes typically of only about 10 14 size 12{"10" rSup { size 8{ - "14"} } } {} s, an indication of the great strength of the forces pulling the nucleons to lower states. The γ size 12{γ} {} decay equation is simply

Z A X N * Z A X N + γ 1 + γ 2 + ( γ decay ) size 12{"" lSub { size 8{Z} } lSup { size 8{A} } X rSub { size 8{N} } rSup { size 8{*} } rightarrow "" lSub { size 8{Z} } lSup { size 8{A} } X rSub { size 8{N} } +γ rSub { size 8{1} } +γ rSub { size 8{2} } + dotsaxis ``` \( γ`"decay" \) } {}

where the asterisk indicates the nucleus is in an excited state. There may be one or more γ s emitted, depending on how the nuclide de-excites. In radioactive decay, γ emission is common and is preceded by γ or β size 12{β} {} decay. For example, when 60 Co β size 12{β rSup { size 8{ - {}} } } {} decays, it most often leaves the daughter nucleus in an excited state, written 60 Ni* . Then the nickel nucleus quickly γ size 12{γ} {} decays by the emission of two penetrating γ size 12{γ} {} s:

60 Ni* 60 Ni + γ 1 + γ 2 . size 12{"" lSup { size 8{"60"} } "Ni" rSup { size 8{*} } rightarrow "" lSup { size 8{"60"} } "Ni"+γ rSub { size 8{1} } +γ rSub { size 8{2} } } {}

These are called cobalt γ size 12{γ} {} rays, although they come from nickel—they are used for cancer therapy, for example. It is again constructive to verify the conservation laws for gamma decay. Finally, since γ size 12{γ} {} decay does not change the nuclide to another species, it is not prominently featured in charts of decay series, such as that in [link] .

There are other types of nuclear decay, but they occur less commonly than α , β , and γ size 12{γ} {} decay. Spontaneous fission is the most important of the other forms of nuclear decay because of its applications in nuclear power and weapons. It is covered in the next chapter.

Test prep for ap courses

A nucleus in an excited state undergoes γ decay, losing 1.33 MeV when emitting a γ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeq4SdCgaaa@379B@ ray. In order to conserve energy in the reaction, what frequency must the γ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeq4SdCgaaa@379B@ ray have?

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95 241 Am MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0baaSqaaiaaiMdacaaI1aaabaGaaGOmaiaaisdacaaIXaaaaOGaaeyqaiaab2gaaaa@3B96@ is commonly used in smoke detectors because its α decay process provides a useful tool for detecting the presence of smoke particles. When 95 241 Am MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0baaSqaaiaaiMdacaaI1aaabaGaaGOmaiaaisdacaaIXaaaaOGaaeyqaiaab2gaaaa@3B96@ undergoes α decay, what is the resulting nucleus? If 95 241 Am MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0baaSqaaiaaiMdacaaI1aaabaGaaGOmaiaaisdacaaIXaaaaOGaaeyqaiaab2gaaaa@3B96@ were to undergo β decay, what would be the resulting nucleus? Explain each answer.

When 95 241 Am MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0baaSqaaiaaiMdacaaI1aaabaGaaGOmaiaaisdacaaIXaaaaOGaaeyqaiaab2gaaaa@3B96@ undergoes α decay, it loses 2 neutrons and 2 protons. The resulting nucleus is therefore 93 237 Np MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0baaSqaaiaaiMdacaaIZaaabaGaaGOmaiaaiodacaaI3aaaaOGaaeOtaiaabchaaaa@3BA9@ .

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For β decay, the nucleus releases a negative charge. In order for charge to be conserved overall, the nucleus must gain a positive charge, increasing its atomic number by 1, resulting in 96 241 Cm. MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0baaSqaaiaaiMdacaaI2aaabaGaaGOmaiaaisdacaaIXaaaaOGaae4qaiaab2gaaaa@3B99@

A 6 14 C MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0baaSqaaiaaiAdaaeaacaaIXaGaaGinaaaakiaaboeaaaa@392A@ nucleus undergoes a decay process, and the resulting nucleus is 7 14 N MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0baaSqaaiaaiEdaaeaacaaIXaGaaGinaaaakiaab6eaaaa@3936@ . What is the value of the charge released by the original nucleus?

  1. +1
  2. 0
  3. -1
  4. -2
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Explain why the overall charge of the nucleus is increased by +1 during the β decay process.

During this process, the nucleus emits a particle with -1 charge. In order for the overall charge of the system to remain constant, the charge of the nucleus must therefore increase by +1.

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Identify the missing particle based upon conservation principles:

N 7 14 + H 2 4 e X + O 8 17 MathType@MTEF@5@5@+=feaagyart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0raaSqaaiaaiEdaaeaacaaIXaGaaGinaaaakiaab6eacqGHRaWkdaqhbaWcbaGaaGOmaaqaaiaaisdaaaGccaqGibGaaeyzaiabgkziUkaadIfacqGHRaWkdaqhbaWcbaGaaGioaaqaaiaaigdacaaI3aaaaOGaae4taaaa@4471@

  1. H 1 2 MathType@MTEF@5@5@+=feaagyart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0raaSqaaiaaigdaaeaacaaIYaaaaOGaaeisaaaa@386D@
  2. H 1 1 MathType@MTEF@5@5@+=feaagyart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0raaSqaaiaaigdaaeaacaaIXaaaaOGaaeisaaaa@386C@
  3. C 6 12 MathType@MTEF@5@5@+=feaagyart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0raaSqaaiaaiAdaaeaacaaIXaGaaGOmaaaakiaaboeaaaa@3928@
  4. C 6 12 MathType@MTEF@5@5@+=feaagyart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0raaSqaaiaaiAdaaeaacaaIXaGaaGOmaaaakiaaboeaaaa@3928@
  5. B 4 8 e MathType@MTEF@5@5@+=feaagyart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0raaSqaaiaaisdaaeaacaaI4aaaaOGaaeOqaiaabwgaaaa@3958@
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Are the following reactions possible? For each, explain why or why not.

  1. U 92 238 R 88 234 a + H 2 4 e
  2. R 88 223 a P 82 209 b + C 6 14
  3. C 6 14 N 7 14 + e + v ¯ e MathType@MTEF@5@5@+=feaagyart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0raaSqaaiaaiAdaaeaacaaIXaGaaGinaaaakiaaboeacqGHsgIRdaqhbaWcbaGaaG4naaqaaiaaigdacaaI0aaaaOGaaeOtaiabgUcaRiaadwgadaahaaWcbeqaaiabgkHiTaaakiabgUcaRiqadAhagaqeamaaBaaaleaacaqGLbaabeaaaaa@4453@
  4. M 12 24 g N 11 23 a + e + + v e MathType@MTEF@5@5@+=feaagyart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0raaSqaaiaaigdacaaIYaaabaGaaGOmaiaaisdaaaGccaqGnbGaae4zaiabgkziUoaaDeaaleaacaaIXaGaaGymaaqaaiaaikdacaaIZaaaaOGaaeOtaiaabggacqGHRaWkcaWGLbWaaWbaaSqabeaacqGHRaWkaaGccqGHRaWkcaWG2bWaaSbaaSqaaiaabwgaaeqaaaaa@4775@
  1. No. Nucleon number is conserved (238 = 234 + 4), but the atomic number or charge is NOT conserved (92 ≠ 88+2).
  2. Yes. Nucleon number is conserved (223 = 209 + 14), and atomic number is conserved (88 = 82 + 6).
  3. Yes. Nucleon number is conserved (14 = 14), and charge is conserved if the electron’s charge is properly counted (6 = 7 + (-1)).
  4. No. Nucleon number is not conserved (24 ≠ 23). The positron released counts as a charge to conserve charge, but it doesn’t count as a nucleon.
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Questions & Answers

how does Neisseria cause meningitis
Nyibol Reply
what is microbiologist
Muhammad Reply
what is errata
Muhammad
is the branch of biology that deals with the study of microorganisms.
Ntefuni Reply
What is microbiology
Mercy Reply
studies of microbes
Louisiaste
when we takee the specimen which lumbar,spin,
Ziyad Reply
How bacteria create energy to survive?
Muhamad Reply
Bacteria doesn't produce energy they are dependent upon their substrate in case of lack of nutrients they are able to make spores which helps them to sustain in harsh environments
_Adnan
But not all bacteria make spores, l mean Eukaryotic cells have Mitochondria which acts as powerhouse for them, since bacteria don't have it, what is the substitution for it?
Muhamad
they make spores
Louisiaste
what is sporadic nd endemic, epidemic
Aminu Reply
the significance of food webs for disease transmission
Abreham
food webs brings about an infection as an individual depends on number of diseased foods or carriers dully.
Mark
explain assimilatory nitrate reduction
Esinniobiwa Reply
Assimilatory nitrate reduction is a process that occurs in some microorganisms, such as bacteria and archaea, in which nitrate (NO3-) is reduced to nitrite (NO2-), and then further reduced to ammonia (NH3).
Elkana
This process is called assimilatory nitrate reduction because the nitrogen that is produced is incorporated in the cells of microorganisms where it can be used in the synthesis of amino acids and other nitrogen products
Elkana
Examples of thermophilic organisms
Shu Reply
Give Examples of thermophilic organisms
Shu
advantages of normal Flora to the host
Micheal Reply
Prevent foreign microbes to the host
Abubakar
they provide healthier benefits to their hosts
ayesha
They are friends to host only when Host immune system is strong and become enemies when the host immune system is weakened . very bad relationship!
Mark
what is cell
faisal Reply
cell is the smallest unit of life
Fauziya
cell is the smallest unit of life
Akanni
ok
Innocent
cell is the structural and functional unit of life
Hasan
is the fundamental units of Life
Musa
what are emergency diseases
Micheal Reply
There are nothing like emergency disease but there are some common medical emergency which can occur simultaneously like Bleeding,heart attack,Breathing difficulties,severe pain heart stock.Hope you will get my point .Have a nice day ❣️
_Adnan
define infection ,prevention and control
Innocent
I think infection prevention and control is the avoidance of all things we do that gives out break of infections and promotion of health practices that promote life
Lubega
Heyy Lubega hussein where are u from?
_Adnan
en français
Adama
which site have a normal flora
ESTHER Reply
Many sites of the body have it Skin Nasal cavity Oral cavity Gastro intestinal tract
Safaa
skin
Asiina
skin,Oral,Nasal,GIt
Sadik
How can Commensal can Bacteria change into pathogen?
Sadik
How can Commensal Bacteria change into pathogen?
Sadik
all
Tesfaye
by fussion
Asiina
what are the advantages of normal Flora to the host
Micheal
what are the ways of control and prevention of nosocomial infection in the hospital
Micheal
what is inflammation
Shelly Reply
part of a tissue or an organ being wounded or bruised.
Wilfred
what term is used to name and classify microorganisms?
Micheal Reply
Binomial nomenclature
adeolu
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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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