Question:
A typical atom has a diameter of about $1.0 \times 10^{-10} \textrm{ m}$.
- What is this in inches?
- Approximately how many atoms are along a 1.0-cm line, assuming they just touch?
Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014. Reprinted by permission of Pearson Education Inc., New York.
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Quick Answer:
a) $3.9 \times 10^{-9} \textrm{ in}$
b) $1.0 \times 10^8 \textrm{ atoms/line}$
Giancoli 7th Edition, Chapter 1, Problem 17
(1:30)
Comments
could you please tell me how you got to 1.0*10^8? looking the problem you laid out... i see 1.0*10^-10/100.... which does not equal your solution?
Hi harringtonwarriorfitness, the work in the video isn't quite how you wrote it in your comment. The problem mentions that "a typical atom has a diameter of about $1.0 \times 10^{-10} \textrm{ m}$". This can be rephrased as "one atom per $1.0 \times 10^{-10} \textrm{ m}$". That's maybe an odd way to say it since it might be more typical to say "$1.0 \times 10^{-10} \textrm{ m}$ per atom", but either way works, and mentioning the atom first puts that in the numerator of the fraction. The "one atom per $1.0 \times 10^{-10} \textrm{ m}$" can be written as the fraction $\dfrac{1 \textrm{ atom}}{1.0 \times 10^{-10} \textrm{ m}}$ and it's this fraction which was missed in your working "1.0*10^-10/100". The "1.0 atom" is on top, and it's divided by "10^-10".
Hope that helps,
Mr. Dychko