Phys18

Phys18 - Unit 18 18.1 18.2 18.3 18.4 18.5 18.6 18.7 Early...

This preview shows pages 1–4. Sign up to view the full content.

1 Unit 18 Early Quantum Physics 18.1 Quantization 18.2 Photoelectric effect 18.3 Compton scattering 18.4 Bohr model 18.5 Wave-particle duality 18.6 Matter waves 18.7 Uncertainty principle 18.1 Quantization A quantity is quantized when its possible values are limited to a discrete set. This is contrast to classical physics, in which most quantities are continuous. Standing values provide an example of quantization in classical physics. The frequency of a standing wave on a string fixed at both ends is quantized. The frequencies of a vibrating string are integral multiples of the fundamental frequency. In modern physics, we realize that electromagnetic radiations occupy energy in discrete amounts of quanta E 0 , where 0 Eh f = , h = 6.626 × 10 34 Js is the Planck’s constant and f is the frequency of the radiation. According to Einstein a blackbody can emit or absorb only a whole number of photon. Example Find the energy of a photon of visible red light of wavelength 670 nm and compare it with the energy of an x-ray photon with frequency 1.0 × 10 19 Hz. Answer: The product of Planck’s constant with each frequency gives the corresponding photon energy. For the 670-nm photon, the frequency and wavelength are related by c = f λ .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2 The frequency of the red light is f=c/ λ . To find the energy we multiply the frequency by Planck’s constant. 34 8 19 9 6.626 10 J s 3.00 10 m/s 3.0 10 J 670 10 m hc Eh f ×⋅ × × === × For the x-ray photon, 34 19 15 6.626 10 J s 1.0 10 Hz = 6.6 10 J f −− = =×⋅ × × The energy of the x-ray photon is more than 20,000 times the energy of a photon of red light. Example A laser produces a beam of light 2.0 mm in diameter. The wavelength is 532 nm and the output power is 20.0 mW. How many photons does the laser emit per second? Answer: The photons all have the same energy since the beam has a single wavelength. The output power is the energy output per unit time. Then the energy output per second is the energy of each photon times the number of photons emitted per second. Energy per second = Energy per photon × photons per second Since c = f , the energy of a photon of wavelength is hc f == . The energy of each photon emitted by the laser is 34 8 19 9 6.626 10 J s 3.00 10 m/s 3.736 10 J 532 10 m E × × × × The number of photons emitted per second is 16 19 0.0200 J s 5.35 10 photons/s 3.736 10 J/photon energy per second Photon per second energy per photon = × × 18.2 Photoelectric effect In the photoelectric effect experiment, the amount of energy that must be supplied to break the bond between a metal and one of its electrons is called the work function φ . Each metal has its own characteristic work function. According to Einstein, EM radiation itself is quantized. The quantum of EM radiation – that is, the smallest indivisible unit – is now called photon. If the photon energy, E = hf , is at least equal
3 to the work function, then absorption of a photon can eject an electron. If the photon energy is greater than the work function, some or all of the extra energy can appear as the ejected electron’s kinetic energy.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 09/06/2010 for the course BSC PHY1417 taught by Professor Prof during the Spring '08 term at HKU.

Page1 / 14

Phys18 - Unit 18 18.1 18.2 18.3 18.4 18.5 18.6 18.7 Early...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online