Chapter 33

# Chapter 33 - CHAPTER Properties of Light 33 1 Why is helium...

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CHAPTER 33 Properties of Light 1* Why is helium needed in a helium–neon laser? Why not just use neon? The population inversion between the state E 2,Ne and the state 1.96 eV below it (see Figure 33-9) is achieved by inelastic collisions between neon atoms and helium atoms excited to the state E 2,He . 2 ∙∙ When a beam of visible white light passes through a gas of atomic hydrogen and is viewed with a spectroscope, dark lines are observed at the wavelengths of the emission series. The atoms that participate in the resonance absorption then emit this same wavelength light as they return to the ground state. Explain why the observed spectrum nevertheless exhibits pronounced dark lines. Although the excited atoms emit the light of the same frequency on returning to the ground state, the light is emitted in a random direction, not exclusively in the direction of the incident beam. Consequently, the beam intensity is greatly diminished. 3 A pulse from a ruby laser has an average power of 10 MW and lasts 1.5 ns. ( a ) What is the total energy of the pulse? ( b ) How many photons are emitted in this pulse? ( a ) E = P t ( b ) E photon = hc / λ (Equ. 17.1); N = E / hc E = 15 mJ N = 5.24 × 10 16 4 A helium–neon laser emits light of wavelength 632.8 nm and has a power output of 4 mW. How many photons are emitted per second by this laser? E photon = hc / ; n = P / hc n = 1.27 × 10 16 s –1 5* The first excited state of an atom of a gas is 2.85 eV above the ground state. ( a ) What is the wavelength of radiation for resonance absorption? ( b ) If the gas is irradiated with monochromatic light of 320 nm wavelength, what is the wavelength of the Raman scattered light? ( a ) Use Equ. 33-2 ( b ) E Raman = E inc E ; Raman = 1240/ E Raman = 1240/2.85 nm = 435 nm E Raman = (1240/320 2.85) eV = 1.025 eV; Raman = 1210 nm 6 ∙∙ A gas is irradiated with monochromatic ultraviolet light of 368 nm wavelength. Scattered light of the same wavelength and of 658 nm wavelength is observed. Assuming that the gas atoms were in their ground state prior to irradiation, find the energy difference between the ground state and the atomic state excited by the irradiation. E = hc / ; use Equs. 17-1 and 17-5 E = 1240/368 eV = 3.37 eV

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Chapter 33 Properties of Light 7 ∙∙ Sodium has excited states 2.11 eV, 3.2 eV, and 4.35 eV above the ground state. ( a ) What is the maximum wavelength of radiation that will result in resonance fluorescence? What is the wavelength of the fluorescent radiation? ( b ) What wavelength will result in excitation of the state 4.35 eV above the ground state? If that state is excited, what are the possible wavelengths of resonance fluorescence that might be observed? ( a ) Excite to 3.2 eV level; decay to 2.11 eV level and ground state; use Equs. 17-1 and 17-5 (b) See the energy level diagram below; λ = hc / E For excitation, max = (1240/3.2) nm = 387.5 nm; fluorescence wavelengths: = 1138 nm and 587.7 nm For excitation, = 1240/4.35 nm = 285 nm The fluorescent wavelengths are as follows: 32 = 1078 nm 21 = 1138 nm 10 = 587.7 nm 31 = 553.6 nm 20 = 387.5 nm 8 ∙∙ Singly ionized helium is a hydrogen-like atom with a nuclear charge of 2 e . Its energy levels are given by E n = –4 E 0 / n 2 , where E 0 = 13.6 eV. If a beam of visible white light is sent through a gas of singly ionized helium, at
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Chapter 33 - CHAPTER Properties of Light 33 1 Why is helium...

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