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College Physics for AP® Courses

Test Prep for AP® Courses

College Physics for AP® CoursesTest Prep for AP® Courses

31.1 Nuclear Radioactivity

1.

A nucleus is observed to emit a γ γ ray with a frequency of 6.3× 10 19  Hz. 6.3× 10 19  Hz. What must happen to the nucleus as a consequence?

  1. The nucleus must gain 0.26 MeV.
  2. The nucleus must also emit an α particle of energy 0.26 MeV in the opposite direction.
  3. The nucleus must lose 0.26 MeV.
  4. The nucleus must also emit a β particle of energy 0.26 MeV in the opposite direction.
2.

A uranium nucleus emits an α particle. Assuming charge is conserved, the resulting nucleus must be

  1. thorium
  2. plutonium
  3. radium
  4. curium

31.3 Substructure of the Nucleus

3.

A typical carbon nucleus contains 6 neutrons and 6 protons. The 6 protons are all positively charged and in very close proximity, with separations on the order of 10-15 meters, which should result in an enormous repulsive force. What prevents the nucleus from dismantling itself due to the repulsion of the electric force?

  1. The attractive nature of the strong nuclear force overpowers the electric force.
  2. The weak nuclear force barely offsets the electric force.
  3. Magnetic forces generated by the orbiting electrons create a stable minimum in which the nuclear charged particles reside.
  4. The attractive electric force of the surrounding electrons is equal in all directions and cancels out, leaving no net electric force.

31.4 Nuclear Decay and Conservation Laws

4.

A nucleus in an excited state undergoes γ γ decay, losing 1.33 MeV when emitting a γ 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 γ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeq4SdCgaaa@379B@ ray have?

5.

95 241 Am 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 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 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.

6.

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. 96 241 Cm. MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0baaSqaaiaaiMdacaaI2aaabaGaaGOmaiaaisdacaaIXaaaaOGaae4qaiaab2gaaaa@3B99@

A 6 14 C 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 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
7.

Explain why the overall charge of the nucleus is increased by +1 during the β decay process.

8.

Identify the missing particle based upon conservation principles:

N 7 14 + H 2 4 e→X+ O 8 17 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 H 1 2 MathType@MTEF@5@5@+=feaagyart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0raaSqaaiaaigdaaeaacaaIYaaaaOGaaeisaaaa@386D@
  2. H 1 1 H 1 1 MathType@MTEF@5@5@+=feaagyart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0raaSqaaiaaigdaaeaacaaIXaaaaOGaaeisaaaa@386C@
  3. C 6 12 C 6 12 MathType@MTEF@5@5@+=feaagyart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0raaSqaaiaaiAdaaeaacaaIXaGaaGOmaaaakiaaboeaaaa@3928@
  4. C 6 12 C 6 12 MathType@MTEF@5@5@+=feaagyart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0raaSqaaiaaiAdaaeaacaaIXaGaaGOmaaaakiaaboeaaaa@3928@
  5. B 4 8 e B 4 8 e MathType@MTEF@5@5@+=feaagyart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0raaSqaaiaaisdaaeaacaaI4aaaaOGaaeOqaiaabwgaaaa@3958@
9.

Are the following reactions possible? For each, explain why or why not.

  1. U 92 238 → R 88 234 a+ H 2 4 e U 92 238 → R 88 234 a+ H 2 4 e
  2. R 88 223 a→ P 82 209 b+ C 6 14 R 88 223 a→ P 82 209 b+ C 6 14
  3. C 6 14 → N 7 14 + e − + v ¯ e 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 M 12 24 g→ N 11 23 a+ e + + v e MathType@MTEF@5@5@+=feaagyart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0raaSqaaiaaigdacaaIYaaabaGaaGOmaiaaisdaaaGccaqGnbGaae4zaiabgkziUoaaDeaaleaacaaIXaGaaGymaaqaaiaaikdacaaIZaaaaOGaaeOtaiaabggacqGHRaWkcaWGLbWaaWbaaSqabeaacqGHRaWkaaGccqGHRaWkcaWG2bWaaSbaaSqaaiaabwgaaeqaaaaa@4775@

31.5 Half-Life and Activity

10.

A radioactive sample has N atoms initially. After 3 half-lives have elapsed, how many atoms remain?

  1. N/3
  2. N/6
  3. N/8
  4. N/27
11.

When P 84 215 o P 84 215 o MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0raaSqaaiaaiIdacaaI0aaabaGaaGOmaiaaigdacaaI1aaaaOGaaeiuaiaab+gaaaa@3BA7@ decays, the product is P 82 211 b. P 82 211 b. MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0raaSqaaiaaiIdacaaIYaaabaGaaGOmaiaaigdacaaIXaaaaOGaaeiuaiaabkgaaaa@3B94@ The half-life of this decay process is 1.78 ms. If the initial sample contains 3.4 x 1017 parent nuclei, how many are remaining after 35 ms have elapsed? What kind of decay process is this (alpha, beta, or gamma)?

31.6 Binding Energy

12.

Binding energy is a measure of how much work must be done against nuclear forces in order to disassemble a nucleus into its constituent parts. For example, the amount of energy in order to disassemble H 2 4 e H 2 4 e MathType@MTEF@5@5@+=feaagyart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0raaSqaaiaaikdaaeaacaaI0aaaaOGaaeisaiaabwgaaaa@3958@ into 2 protons and 2 neutrons requires 28.3 MeV of work to be done on the nuclear particles. Describe the force that makes it so difficult to pull a nucleus apart. Would it be accurate to say that the electric force plays a role in the forces within a nucleus? Explain why or why not.

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