Asked by Bella

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.
Answered by bobpursley
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
Answered by Bella
Thank you for your help.
Answered by sabano chmest
1*10^4
Answered by Rich P.
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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