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The Quantum nature of radiation

The Quantum nature of radiation. For years it was accepted that light travels as particles (though with little direct evidence). Largely based on my corpuscular theory of light. Isaac Newton 1642-1727. The Wave theory of radiation.

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The Quantum nature of radiation

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  1. The Quantum nature of radiation For years it was accepted that light travels as particles (though with little direct evidence). Largely based on my corpuscular theory of light Isaac Newton 1642-1727

  2. The Wave theory of radiation However, this idea was overthrown after Young demonstrated beyond doubt the wave nature of light in his double slit experiment (1801).

  3. The photoelectric effect However, around the turn of the 20th century, an effect that became known as the photoelectric effect defied explanation using the wave model. Oh no!

  4. The photo-electric effect. • A clean zinc plate is given a negative charge • Nanocoulomb meter indicates movement of electrons Zinc plate Nano-coulomb meter 0.0

  5. The photo-electric effect. • The nanometer indicated that when visible light shone on the plate there no movement of electrons Visible light Zinc plate Nano-coulomb meter 0.0

  6. The photo-electric effect. • However, replaced with UV light the plate discharged immediately. Ultraviolet light Zinc plate Nano-coulomb meter 124

  7. Photoelectric Effect • This phenomenon is called the photoelectric effect • Photo (light) causing emission of electrons (electric) • Energy from the light is given to the electrons in the zinc plate and some electrons near the surface of the zinc gain enough energy to escape from the attraction of the positive charge of the nucleus.

  8. Photon electrons

  9. As visible light is more intense (has more energy) than UV, how come it only happens with UV light?

  10. ELECTROMAGNETIC SPECTRUM 740nm 370nm

  11. Interesting features 1. For every metal, there is a certain frequency of light (the threshold frequency), below which no electrons are emitted, no matter how intense the light. K.E. of emitted electrons Frequency Threshold frequency (fo)

  12. Interesting features 2. The number of photoelectrons emitted (above the threshold frequency) only depends on the intensity of light.

  13. It looks like my theory’s buggered! A few problems Using classical physics it could not be explained why there is a threshold frequency below which no electrons are emitted. Surely if the light was intense enough, the electrons would gain enough energy to escape at any frequency? Also, why was there no time delay?

  14. A few problems Classical physics could also not explain why the number of electrons emitted depended only on the intensity and not the frequency. I know someone who may be able to help. Max Planck 1858 - 1947

  15. Einstein to the Rescue!!! He explained this effect in terms of quantum theory. Quantum : fixed indivisible amount Eg/ charge is quantised because all electric charges are made up of a whole number multiple of ‘fixed’ electron charges (1.6x10-19C).

  16. Electromagnetic radiation does exist as continuous waves but rather as discrete bundles (quanta) of energy = PHOTONS Each photon has a discrete amount of energy, hf, where f is the frequency of the radiation and h is a universal constant: Planck constant (h=6.63x10-34Js). Energy = hf This theory successfully explained the distribution of energy with wavelength for the radiation from a hot body. Einstein applied Planck’s theory to the photoelectric effect.

  17. Instead of the energy from the light being gradually absorbed by the electrons near the surface of the zinc until they had enough energy to escape… Einstein reasoned that the electron would be emitted only if a single quantum of light had enough energy for the electron to escape. Frequency of Visible Light not high enough for a photon to provide the energy for an electron to escape…… BUT UV light does!

  18. ELECTROMAGNETIC SPECTRUM UV has a higher freq. than visible light. UV quanta each have sufficient energy to release an electron. So zinc plate discharges immediately in the presence of UV. Intensity depends on : no. quanta energy associated with each quanta

  19. - - - - - - - Photons Free electrons Metal Here, the photon energy is less than the minimum required for the electrons to escape – no electrons are produced

  20. - - - - - - - - photoelectron Photons Free electrons Metal The photon energy is greater than the minimum required for the electrons to escape – photoelectrons are produced at a range of kinetic energies up to a maximum value.

  21. 1. Blue light has a frequency of 7.7x1012Hz, while red light has a frequency of 4.3x1014 Hz. Calculate the energy of a photon of each. Planck Constant h=6.6x10-34 Js • Calculate the energy of: • Radio waves of frequency 100MHz • Microwaves of wavelength 10cm • Infra red waves of wavelength 1mm • X-rays of wavelength 10nm • Speed of light = 3.0x108m/s • Planck constant = h=6.6x10-34 Js Remember: f = c/λ

  22. Question: • An advertisement for a green laser pointer is given below: • What is the efficiency of the laser in converting electrical energy to light energy? • What is the intensity (radiation flux density) of the light beam close to the laser? • How much energy is in one photon of the light? • How many photons per second are emitted by the laser? • The manufacture claims that this laser is ‘much brighter’ than a 5mW red laser. Suggest why this statement may be justified.

  23. The graph shows the energy of the photo-electrons that are emitted from Li metal for different frequencies of light. Light of a f less than 5.6x1014 Hz doesn’t give any photo-electrons. Threshold Frequency, f0 The graph goes up in a straight line as the f of light goes up, the energy of the emitted electrons goes up as well. The gradient of the line is Planck constant, h. This is true of all metals.

  24. Using E=hf we can convert f of the incident light to energy of the photons. Graph of KE of emitted electrons Vs Photon Energy. When the photon energy is 3eV the energy emitted is 0.7eV, that’s 2.3eV less! B/c graph is straight line, this difference of 2.3eV is true for all photo-electrons emitted frm Li. Where does the graph cut the x-axis?

  25. Frm the graph 2.3eV is the smallest quantity of energy that is needed for electron to be given off for Li(s). It is called the work function of this metal; Фo Work function,Фo = hf So, some of the photons energy is needed to remove the electron (Фo), and any surplus becomes kinetic energy. Energy of photon energy required to remove electron + KEmax of ejected electron E = hf = Фo + ½m(vmax)2

  26. Nb/ The work function is larger for less reactive metals, like zinc. It is harder to remove electrons from these metals, and so ultra violet light is needed. In qu. You might be given the cut off wavelength, λ0 , the longest wavelength of light that emits photo-electrons. You can then use this to fine the threshold frequency.

  27. Qu.The cut off wavelength, λ0 for potassium is 550nm. Calculate the work function in joules. Qu.The work function for zinc is 4.3eV. Explain why photoelectric emission is observed when ultraviolet light of wavelength in the order of 200nm is shone onto a zinc plate but not when a 60W filament lamp is used.

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