Explanation Figure P42.58 gives the energy level transition diagram for He − Ne laser. Electrical excitation given to Helium atom excites an electron from the Helium atom to its excited state of energy 20.61 eV . This high energy electron excites the ground level Neon atom to excited state shown as E 3 ∗ . Lasing action of the laser takes place when the electron from E 3 ∗ falls down to state E 2 . When electron Write the expression for the energy of the photon released when electron drops from E ∗ to E 2 . E p = h c λ (I) Here, E p is the energy of the photon released when electron drops from E 3 ∗ to E 2 , h is the Planck’s constant, c is the speed of light, and λ is the wavelength of photon emitted. Write the expression to find E 3 ∗ . E 3 ∗ = E 2 + E p (II) Rewrite expression (II) to find E p . E p = E 3 ∗ − E 2 (III) Use expression (I) in (III)

Physics for Scientists and Engineers With Modern Physics

9th Edition

ISBN: 9781133953982

Author: SERWAY, Raymond A./

Publisher: Cengage Learning

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Chapter 42, Problem 58P

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Show that the wavelength of the red He−Ne laser light is 633 nm.

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Chapter 42, Problem 57P

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Chapter 42 Solutions

Physics for Scientists and Engineers With Modern Physics

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Show that the wavelength of the red He − Ne laser light is 633 nm .

Solution Summary: The author illustrates the energy level transition diagram for the red He- Ne laser. The electron excites the ground level Neon atom to excited state.

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Show that the wavelength of the red He−Ne laser light is 633 nm.

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Chapter 42 Solutions