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marihelenh

  • one year ago

The gold foil used by Ernst Rutherford in his investigations 1 um thick. Given that the radius of gold is approximately 130 pm, how many atoms thick was the foil?

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  1. aaronq
    • one year ago
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    How many atoms fit across the foil (assuming they're liked up side by side) \(1~\mu m=1.0*10^{6}~pm\)

  2. marihelenh
    • one year ago
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    \[\frac{ 130p m }{ 1 } * \frac{ 1\mu m }{ 1.0*10^{6}p m }\]

  3. marihelenh
    • one year ago
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    Is this how I would set that part up?

  4. marihelenh
    • one year ago
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    @aaronq

  5. aaronq
    • one year ago
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    yes that would give you the size of one atom in micrometers, then you divide 1 micrometer (the thickness of the foil) by the size of 1 atom (in micrometers)

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