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                                                                                                          5
                                                                    Chapter 7 Quantum Physics   Physics
                      Example 1                                   6.  The relationship between the momentum of
                                                                    particle, p and its wavelength, λ is
                  Calculate the photon energy for light with
                  wavelengths 450 nm and 700 nm. Compare both                       λ = —–
                                                                                        h
                  photons.                                                              p
                   Solution                                         which h is Planck’s constant (6.63 × 10  J s)
                                                                                                  –34
                  Planck’s constant, h = 6.63 × 10  J s
                                           –34
                  Speed of light in vacuum, c = 3.00 × 10  m s
                                                     –1
                                                 8
                  Wavelength, λ = 450 × 10  m
                                       –9
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                  Wavelength, λ  = 700 × 10  m
                                       –9
                              2                                                              2
                                                                                            p
                                          hc
                                                                                     (mv)
                                                                      1
                                                                            m
                                                                                        2
                                   E = hf = ——                     K = —mv  × —– ➞  K = ——— = ——
                                                                         2
                                           λ                          2     m        2m    2m
                  Photon energy of 450 nm:                         which p = mv, p = ABBBB
                                                                                 2mK
                                        8
                               1
                  E  = 6.63 × 10 –34 3.00 × 10 –9 2
                                —————–
                    1            450 × 10
                    = 4.42 × 10  J                                7.  The larger the momentum of particle, the shorter
                              –19
                  Photon energy of 700 nm:                          the wavelength produced. The momentum of
                                                                                       p
                                                                                        2
                                        8
                               1
                  E  = 6.63 × 10 –34 3.00 × 10 –9 2                 particle is p = mv, then —— = K
                                —————–
                    2            700 × 10                                              2m
                    = 2.84 × 10  J
                              –19
                                                                                    h
                                                                                           h
                  The shorter the wavelength of light, the higher              λ = —–– = ————
                                                                                   mv
                                                                                        ABBBB
                  the photon energy.                                                      2mK
                                                                    where m is the mass of particle, v is the velocity
                                                                    of particle and K is the kinetic energy of particle.
                  Wave-Particle Duality                           8.  Since the value of  h  is  too  small, the particle
                                                                    with large mass will have too short of de Broglie
                                                                    wavelength to  be  detected.  Thus,  the  wave
                                                                    characteristics cannot be observed.
                   Wavelength simulation                          9.  In 1927, the presence of wave properties of
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                                                                    electrons was confirmed through the electron
                                                                    diffraction experiments.
                   1.  Light has wave properties because it shows the    10.  Photograph 7.1 shows the diffraction pattern of
                     phenomena of diffraction and interference.     electrons through a thin layer of graphite. The
                                                                    pattern in Photograph 7.2 resembles the light
                   2.  Object has particle properties because it has   diffraction pattern through an aperture.
                     momentum, kinetic energy and also collide with
                     each other.
                   3.  Louis de Broglie introduced a hypothesis states
                     that all particles can exhibit wave characteristics
                     in year 1924.
                   4.  However, it is experimentally difficult to show
                     the wave characteristics of particles with large
                     masses.
                   5.  Thus, Louis de Broglie predicted that the wave      Photograph 7.1   Photograph 7.2
                     characteristics can be shown by light particles    11.  This observation proved de Broglie’s hypothesis.   CHAP
                     such as electron, proton and neutron.                                                  7
                 7.1.3                                                                                477
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