Integrated Science
Integrated Science
7th Edition
ISBN: 9780077862602
Author: Tillery, Bill W.
Publisher: Mcgraw-hill,
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Chapter 7, Problem 9PEB
To determine

The amount of energy of a photon of ultraviolet light greater than the energy of an average photon of visible light.

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Answer to Problem 9PEB

The energy of a photon of ultraviolet light is 3.08×1019J more than the energy of an average photon of visible light.

Explanation of Solution

Write the expression for the energy of a photon.

    E=hf                                                                                   (I)

Here, E is the energy of a photon h is the Planck's constant and f is the frequency.

Write the expression for the speed of light.

    c=λff=cλ

Here, c is the speed of light and λ is the wavelength of light.

Substitute cλ for f in equation (I) to find E.

    E=hcλ                                                                               (II)

Write the expression for the difference in energy of ultraviolet light and the visible light.

    ΔE=EUVEvisible                                                              (III)

Here, EUV is the energy of ultraviolet light and Evisible is the energy of visible light.

Conclusion:

Substitute 6.63×1034Js for h, 3.0×108m/s for c and 3.00×107m for λ in equation (II) to find EUV.

    EUV=(6.63×1034Js)(3.0×108m/s)3.00×107m=6.63×1019J

Substitute 6.63×1034Js for h, 3.0×108m/s for c and 5.60×107m for λ in equation (II) to find Evisible.

    Evisible=(6.63×1034Js)(3.0×108m/s)5.60×107m=3.55×1019J

Substitute 6.63×1019J for EUV and 3.55×1019J for Evisible in equation (III) to find ΔE.

    ΔE=6.63×1019J3.55×1019J=3.08×1019J

Therefore, the energy of a photon of ultraviolet light is 3.08×1019J more than the energy of an average photon of visible light.

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