APPLE

1. Operation: Annihilate! KABOOM!!! ANTI-APPLE APPLE

2. Note to the teacher: This lesson and its companion lesson “ENERGY’S SPAWN” deal with the transformation of matter to energy (and energy to matter) using the formula E = mc2 it is recommended that you go through “CONSERVATION OF ENERGY-REVISTED” in the Background Materials with your students before embarking upon these two lessons:

3. 4 PARTICLES… electron neutrino up quark down quark

4. 4 PARTICLES… 4 ANTIPARTICLES… electron neutrino up quark down quark positron anti-neutrino anti-down quark anti-up quark

5. 4 PARTICLES… 4 ANTIPARTICLES… electron neutrino up quark down quark positron anti-neutrino anti-down quark anti-up quark WILL THERE BE ANYTHING ELSE WITH THAT, PARTICLE-WISE?

6. YES, LOTS.

7. YES, LOTS. BUT FOR NOW, WE WILL INTRODUCE YOU TO JUST ONE MORE…

8. YES, LOTS. BUT FOR NOW, WE WILL INTRODUCE YOU TO JUST ONE MORE… THE PHOTON (the symbol for the photon is )

9. WE SHALL DISPENSE WITH THE ARMS AND LEGS FROM HERE ON, AND JUST REPRESENT IT LIKE THIS: THE PHOTON () The photon can be considered as a tiny packet of energy. For example, Light is a stream of photons.

10. THE ENERGY OF A SINGLE PHOTON IS GIVEN BY THE FOLLOWING SIMPLE LITTLE FORMULA: E = hf* We will come back to this formula later, but first… *You will find a short tutorial on this formula in “Oh No!! Formulas!” in the Background Materials:

11. PREVIOUSLY, WE SAW WAYS IN WHICH MATTER AND ANTIMATTER ARE SIMILAR AND WAYS IN WHICH THEY DIFFER. INTERESTING THOUGH THAT MAY BE, IT DOESN’T REALLY EXPLAIN WHY THERE IS SO MUCH OF A FUSS ABOUT ANTIMATTER. TO GET AN IDEA OF WHAT WE ARE TALKING ABOUT, HAVE A LOOK AT THE FOLLOWING CLIP...

12. From the film “Angels & Demons” (2009) To see the video, click on the picture

13. Who wrote this stuff???

14. I’m going to have a little chat with my agent…

15. NOW, LET’S TAKE A CLOSER LOOK AT SOME OF THAT RIVETING DIALOGUE… (THE INTERESTING BITS ARE IN BLUE)

16. Vittoria Vetra: That canister contains an extremely combustible substance called antimatter. We need to locate it immediately or evacuate Vatican City! Cdr. Richter: I’m quite familiar with incendiaries, Miss Vetra. I’m unaware of antimatter being used as such. Vittoria Vetra: Well, it’s never been generated in sufficient quantities before. It’s a way of studying the origins of the Universe... …The antimatter is suspended there, in an airtight, nano-composite shell with electromagnets on each end. Butif it were to fall out of suspension and come in contact with matter,say at the bottom of the canister, then the two opposing forces would annihilate one another, violently. Cdr. Richter: And what might cause it to fall out of suspension? Vittoria Vetra: The battery going dead. Which it will—just before midnight. Cdr. Richter: What kind of annihilation? How violent? Vittoria Vetra: A cataclysmic event. A blinding explosion equivalent to about five kilotons. Robert Langdon: Vatican City will be consumed by light.

17. “IF [ANTIMATTER] WERE TO COME INTO CONTACT WITH MATTER THEY WOULD ANNIHILATE EACH OTHER—VIOLENTLY.”

18. “IF [ANTIMATTER] WERE TO COME INTO CONTACT WITH MATTER THEY WOULD ANNIHILATE EACH OTHER—VIOLENTLY.” TRUE.

19. “IF [ANTIMATTER] WERE TO COME INTO CONTACT WITH MATTER THEY WOULD ANNIHILATE EACH OTHER—VIOLENTLY.” TRUE. ANTIMATTER + MATTER = ENERGY

20. ANTIMATTER + MATTER = ENERGY LET’S LOOK AT THIS MORE CLOSELY FOR THE CASE OF AN ELECTRON AND A POSITRON:

21. ANTIMATTER + MATTER = ENERGY LET’S LOOK AT THIS MORE CLOSELY FOR THE CASE OF AN ELECTRON AND A POSITRON: MATTER electron

22. ANTIMATTER + MATTER = ENERGY LET’S LOOK AT THIS MORE CLOSELY FOR THE CASE OF AN ELECTRON AND A POSITRON: + MATTER ANTIMATTER + electron positron

23. ANTIMATTER + MATTER = ENERGY LET’S LOOK AT THIS MORE CLOSELY FOR THE CASE OF AN ELECTRON AND A POSITRON: + = MATTER ANTIMATTER ENERGY two photons + electron positron

24. IT IS VERY SIMPLE TO FIND OUT HOW MUCH ENERGY IS PRODUCED.

25. IT IS VERY SIMPLE TO FIND OUT HOW MUCH ENERGY IS PRODUCED. WE NEED THE MASS OF THE ELECTRON (me- ) AND THE POSITRON (me+) AND THE FORMULA E = mc2

26. IT IS VERY SIMPLE TO FIND OUT HOW MUCH ENERGY IS PRODUCED. WE NEED THE MASS OF THE ELECTRON (me- ) AND THE POSITRON (me+) AND THE FORMULA E = mc2 Let’s do a short calculation…

27. The masses of an electron and a positron are the same: me- = me+ = 9.11 × 10-31 kg So the total mass of both particles is: me- + me+ = m = 1.82 × 10-30 kg and, c = 3.00 × 108 m/s So, E = mc2 = 1.82 × 10-30 × (3.00 × 108)2 E = 1.64 × 10-13 J

28. THIS MEANS THAT THE MASS OF THE ELECTRON AND THE MASS OF THE POSITRON HAVE BEEN CONVERTED INTO TWO LITTLE PACKETS OF ENERGY, THE TWO PHOTONS, EACH OF THEM CONTAINING 0.82 × 10-13 JOULES OF ENERGY: + 9.11 × 10-31 kg 9.11 × 10-31 kg 0.82 × 10-13 J + 0.82 × 10-13 J electron positron

29. TO BE MORE ACCURATE, THE TWO PHOTONS MOVE OFF IN OPPOSITE DIRECTIONS AFTER THE ANNIHILATION*… * The reasons for this go beyond the scope of this module, but it is to satisfy conservation of momentum requirements.

30. TO BE MORE ACCURATE, THE TWO PHOTONS MOVE OFF IN OPPOSITE DIRECTIONS AFTER THE ANNIHILATION*… * The reasons for this go beyond the scope of this module, but it is to satisfy conservation of momentum requirements.

31. TO BE MORE ACCURATE, THE TWO PHOTONS MOVE OFF IN OPPOSITE DIRECTIONS AFTER THE ANNIHILATION*… ANNIHILATION * The reasons for this go beyond the scope of this module, but it is to satisfy conservation of momentum requirements.

32. TO BE MORE ACCURATE, THE TWO PHOTONS MOVE OFF IN OPPOSITE DIRECTIONS AFTER THE ANNIHILATION*… ANNIHILATION * The reasons for this go beyond the scope of this module, but it is to satisfy conservation of momentum requirements.

33. THE CALCULATION BECOMES MUCH SIMPLER IF, INSTEAD OF USING MASS UNITS (KILOGRAMS) AND JOULES, WE USE THE ENERGY UNITS MEGA ELECTRON VOLTS (MeV) FOR ALL OF THE PARTICLES… For information on the electron volt, see “Units!” in the Background Materials:

34. THE CALCULATION BECOMES MUCH SIMPLER IF, INSTEAD OF USING MASS UNITS (KILOGRAMS) AND JOULES, WE USE THE ENERGY UNITS MEGA ELECTRON VOLTS (MeV) FOR ALL OF THE PARTICLES… + 0.51 MeV 0.51 MeV 0.51 MeV + 0.51 MeV electron positron

35. WE CAN EVEN FIND OUT IN WHAT FORM THIS ENERGY IS BY USING E = hf

36. TAKING E = hf and rearranging in terms of f gives: Hz As an exercise, use to show that the wavelength λ of the photon is 2.42 × 10-12 m.

37. Frequencies of 1020 Hz (or wavelengths of 10-12 m) are characteristic of GAMMA RAYS (γ-rays)

38. Frequencies of 1020 Hz (or wavelengths of 10-12 m) are characteristic of GAMMA RAYS (γ-rays) f

39. SO, MATTER AND ANTIMATTER DO INDEED ANNIHILATE EACH OTHER AND THEIR MASSES ARE CONVERTED TO ENERGY.

40. SO, MATTER AND ANTIMATTER DO INDEED ANNIHILATE EACH OTHER AND THEIR MASSES ARE CONVERTED TO ENERGY. AND WE NOW KNOW THAT THE ENERGY PRODUCED WHEN THEY ANNIHILATE EACH OTHER IS IN THE FORM OF GAMMA RAYS:

41. SO, MATTER AND ANTIMATTER DO INDEED ANNIHILATE EACH OTHER AND THEIR MASSES ARE CONVERTED TO ENERGY. AND WE NOW KNOW THAT THE ENERGY PRODUCED WHEN THEY ANNIHILATE EACH OTHER IS IN THE FORM OF GAMMA RAYS: + electron positron two γ-ray photons

42. As an exercise, find out how much energy is produced when a proton is annihilated by an antiproton, given that the mass of a proton (mp) IS 1.67×10-27 kg.

43. As an exercise, find out how much energy is produced when a proton is annihilated by an antiproton, given that the mass of a proton (mp) IS 1.67×10-27 kg. (Answer: 3.01×10-10 J)

44. As an exercise, find out how much energy is produced when a proton is annihilated by an antiproton, given that the mass of a proton (mp) IS 1.67×10-27 kg. (Answer: 3.01×10-10 J) IN WHAT FORM WOULD THIS ENERGY BE?

45. As an exercise, find out how much energy is produced when a proton is annihilated by an antiproton, given that the mass of a proton (mp) IS 1.67×10-27 kg. (Answer: 3.01×10-10 J) IN WHAT FORM WOULD THIS ENERGY BE? (Answer: γ-rays)

46. “A BLINDING EXPLOSION EQUIVALENT TO ABOUT 5 KILOTONS.”

47. “A BLINDING EXPLOSION EQUIVALENT TO ABOUT 5 KILOTONS.” 5 KILOTONS OF WHAT? LET’S ASSUME THAT THEY MEAN 5 KILOTONS OF TNT. THIS LEADS TO AN INTERESTING LITTLE EXERCISE:

48. WHAT Total MASS OF ANTIMATTER + MATTER WOULD HAVE TO ANNIHILATE TO GIVE THE SAME AMOUNT OF ENERGY AS 5 KILOTONS OF TNT?

49. WHAT Total MASS OF ANTIMATTER + MATTER WOULD HAVE TO ANNIHILATE TO GIVE THE SAME AMOUNT OF ENERGY AS 5 KILOTONS OF TNT? When 1 kiloton of TNT explodes, it gives roughly 4.2 × 1012 joules of energy. So 5 kilotons would give 2.1 × 1013 joules. Using E = mc2, calculate the total mass (antimatter + matter) required to obtain 2.1 × 1013 joules of energy.

50. WHAT Total MASS OF ANTIMATTER + MATTER WOULD HAVE TO ANNIHILATE TO GIVE THE SAME AMOUNT OF ENERGY AS 5 KILOTONS OF TNT? When 1 kiloton of TNT explodes, it gives roughly 4.2 × 1012 joules of energy. So 5 kilotons would give 2.1 × 1013 joules. Using E = mc2, calculate the total mass (antimatter + matter) required to obtain 2.1 × 1013 joules of energy. (Answer: 0.00023 kg OR 0.23 g)