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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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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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