quantum and nuclear physics (b)

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

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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. - PowerPoint PPT Presentation

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Page 1: Quantum and  Nuclear  Physics (B)

Quantum and Nuclear Physics (B)

Problem Solving

Mr. KlapholzShaker Heights

High School

Page 2: Quantum and  Nuclear  Physics (B)

Problem 1An 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?

Page 3: Quantum and  Nuclear  Physics (B)

Solution 1Electron-Volts is not an SI unit, so let’s convert:

7.7 x 106 eV ( 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 x 10-19] × [(79)(1.6x10-19)] / R2

R = ?R = 3.0 x 10-14 m

(Memorize size of nucleus ≈ 10-15 m)

Page 4: Quantum and  Nuclear  Physics (B)

Problem 2A sample of a radioactive isotope contains 1.0 x 1024 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.

Page 5: Quantum and  Nuclear  Physics (B)

Solution 2

a) 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.

Page 6: Quantum and  Nuclear  Physics (B)

Solution 2

b) The “initial activity” is the original number of nuclei that decay per second.

A = lNA0 = 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 1019 Bq

Page 7: Quantum and  Nuclear  Physics (B)

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 x 1023 nuclei

Page 8: Quantum and  Nuclear  Physics (B)

Solution 2

d) Convert 30 minutes to SI units:( 30 minutes )( 60 s / minute ) = 1800 sThe number of nuclei that are still not

disintegrated is:N = N0e-lt

N = (1.0 x 1024)e-(3.2 x 10-5)(1800) = ?

N = 9.4 x 1023 nucleiCheck: this is more than N after 12 hours, and

less than N0.

Page 9: Quantum and  Nuclear  Physics (B)

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

Time / s Counts per second

0 210

1 110

2 60

3 35

Page 10: Quantum and  Nuclear  Physics (B)

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

Time / s Counts per second MINUS background

0 200

1 100

2 50

3 25

Page 11: Quantum and  Nuclear  Physics (B)

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.