Posted by Bella on Tuesday, September 13, 2011 at 5:09am.
The gold foil Rutherford used in his scattering experiment had a thickness of approximately 4×10^−3 mm. If a single gold atom has a diameter of 2.9 x10^-8cm , how many atoms thick was Rutherford's foil?
Express your answer using one significant figure.
convert 4x10^-3mm to cm, and you get .04cm.
gold foil thickness divided by atom diameter.
.04cm/2.9x10^-8cm= 1379310 atoms
Is that correct? Also how am I suppose to put that into 1 significant figure? I've already tried 1x10^6 and that was wrong.
Chemistry (urgent) - bobpursley, Tuesday, September 13, 2011 at 5:25am
One significant figure? What is your weigh? 200?100?
Ok, convert 4E-3mm to cm. 4E-3mm*1cm/10mm=.4E-3cm=4E-4cm So I see an error there.
number atoms=4E-4/2.9E-8= 1E4
Chemistry (urgent) - Bella, Tuesday, September 13, 2011 at 5:30am
Thank you for your help.
Chemistry (urgent) - sabano chmest, Friday, September 16, 2011 at 12:03am
Chemistry (urgent) - Rich P., Tuesday, November 6, 2012 at 9:57am
Too late to help you much now, but maybe enough to help someone else. You mis-converted the thickness of the foil by moving the decimal point the wrong direction. 0.004 mm is equal to 0.0004 cm, not 0.04 cm. So your final answer is off by a factor of 100. 1x10^4, is the correct answer to one significant digit.
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