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Quantum and Nuclear Physics (B)

Mr. Klapholz Shaker Heights High School. Quantum and Nuclear Physics (B). Problem Solving. Problem 1. An alpha particle is shot directly at a gold atom with a kinetic energy of 7.7 MeV . How close can the alpha get to the center of the nucleus?. Solution 1.

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Quantum and Nuclear Physics (B)

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  1. Mr. Klapholz Shaker Heights High School Quantum and Nuclear Physics (B) Problem Solving

  2. Problem 1 An alpha particle is shot directly at a gold atom with a kinetic energy of 7.7 MeV. How close can the alpha get to the center of the nucleus?

  3. Solution 1 Electron-Volts is not an SI unit, so let’s convert: 7.7 x 106eV( 1.602 x 10-19 J / 1 eV) = 1.2 x 10-12 J Initial Ek of a = Electrical Ep at closest approach 1.2 X 10-12= kQaQgold / R2 1.2 X 10-12 = [9.0x109] × [1.6 x10-19] × [(79)(1.6x10-19)] / R2 R = ? R = 3.0 x 10-14m (Memorize size of nucleus ≈ 10-15m)

  4. Problem 2 A sample of a radioactive isotope contains 1.0 x 1024 atoms and it has a half life of 6 hours. Find: • The decay constant. • The initial activity. • The number of original nuclei still present after 12 hours. • The number of original nuclei still present after 30 minutes.

  5. Solution 2 • We know that T½ = 6.0 hours. How do we find the decay constant? ln(2) / l= T½ We will need S.I. units. T½ = ? T½ = ( 6 hours )(3600 s / hour) = 21600 s l= ln(2) / T½ = ln(2) / (21600 s) l= 3.2 x 10-5 s-1 This is the probability that any single nucleus will decay, in one second.

  6. Solution 2 b) The “initial activity” is the original number of nuclei that decay per second. A = lN A0= lN0 A0 =[ 3.2 x 10-5 s-1][ 1.0 x 1024] A0 = 3.2 x 1019 nuclei per second A0 = 3.2 x 1019 Becquerel A0 = 3.2 x 1019Bq

  7. Solution 2 c) 12 hours is 2 half lives. So, the number of original nuclei remaining is one-fourth of the original. N= ¼ × [ 1.0 x 1024] N =2.5 x1023 nuclei

  8. Solution 2 d) Convert 30 minutes to SI units: ( 30 minutes )( 60 s / minute ) = 1800 s The number of nuclei that are still not disintegrated is: N = N0e-lt N = (1.0 x1024)e-(3.2 x 10-5)(1800) = ? N =9.4 x1023 nuclei Check: this is more than N after 12 hours, and less than N0.

  9. Problem 3 The background radiation rate is 10 counts per second. Find the half life.

  10. Solution 3 If you subtract the background rate, you’ll see that the half life is one second.

  11. Tonight’s HW: Go through the Quantum and Nuclear section in your textbook and scrutinize the “Example Questions” and solutions. Bring in your questions to tomorrow’s class.

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