The complete question is The diameter of a copper (Cu) atom is roughly 1.3 x 10^-10 m. How many times can you divide evenly a piece of 10-cm copper wire until it is reduced to two separate copper atoms? (Assume there are appropriate tools for this procedure and that copper atoms are lined up in a straight line, in contract with each other.) Solution wt explanation. TIA!
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2 Answers

The width of two Cu atoms is 2.6×10⁻¹⁰m=2.6×10⁻⁸cm.

We need to divide 10cm by this to find out how many times the copper wire needs to be divided:

10÷(2.6×10⁻⁸)=10x10⁸/2.6=3.8462×10⁸=384,620,000 approx.

(If we were to halve the length of copper wire to 5cm, then halve again to 2.5cm, and so on, we would only have to do this between 28 and 29 times to get to the width of two atoms!)

by Top Rated User (1.2m points)
The diameter of a copper atom is roughly  d = 1.3 x10^{-10} m

The length of the copper wire is L = 10  cm = 0.1  m.

To reduce the copper wire into two separate we need to divide it into let say ''n'' equal divisions.
Now, we need to divide the length ''L'' into 2^n times (each time dividing by 2) to obtain the diameter ''d''( which is a single atom).

(L) / (2^n)  = d
(0.1  m) / (2^n)  = 1.3 \times 10^{-10} \ m
 2^n = 769230769
Taking log on both sides we have
log 2^n  = log   769230769
n log  2  =  log  \ 769230769
n = 29 \ times
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