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

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Presentation on theme: "Quantum and Nuclear Physics (B) Problem Solving Mr. Klapholz Shaker Heights High School."— Presentation transcript:

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

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 10 6 eV ( 1.602 x 10 -19 J / 1 eV ) = 1.2 x 10 -12 J Initial E k of  = Electrical E p at closest approach 1.2 X 10 -12 = kQ  Q gold / R 2 1.2 X 10 -12 =  9.0x10 9  1.6 x 10 -19  (79)(1.6x10 -19 )  / R 2 R = ? R = 3.0 x 10 -14 m (Memorize size of nucleus ≈ 10 -15 m)

4 Problem 2 A sample of a radioactive isotope contains 1.0 x 10 24 atoms and it has a half life of 6 hours. Find: a)The decay constant. b)The initial activity. c)The number of original nuclei still present after 12 hours. d)The number of original nuclei still present after 30 minutes.

5 Solution 2 a)We know that T ½ = 6.0 hours. How do we find the decay constant? ln(2) /  = T ½ We will need S.I. units. T ½ = ? T ½ = ( 6 hours )(3600 s / hour) = 21600 s  = ln(2) / T ½ = ln(2) / (21600 s)  = 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 = N A 0 = N 0 A 0 = [ 3.2 x 10 -5 s -1 ] [ 1.0 x 10 24 ] A 0 = 3.2 x 10 19 nuclei per second A 0 = 3.2 x 10 19 Becquerel A 0 = 3.2 x 10 19 Bq

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 10 24 ] N = 2.5 x 10 23 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 = N 0 e - t N = ( 1.0 x 10 24 ) e -(3.2 x 10-5)(1800) = ? N = 9.4 x 10 23 nuclei Check: this is more than N after 12 hours, and less than N 0.

9 Problem 3 The background radiation rate is 10 counts per second. Find the half life. Time / sCounts per second 0210 1110 260 335

10 Solution 3 If you subtract the background rate, you’ll see that the half life is one second. Time / s Counts per second MINUS background 0200 1100 250 325

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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