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I wanted to know if I was thinking about this problem right and set it up correctly. It's been a while since I had to do math like this. The radius of the hydrogen atom is $0.529\times10^{−10}m$ and the radius of the nucleus is $1.2\times10^{−15}m$.

I have a ball bearing with a radius of $1.5mm (1.5\times10^{-3}m)$.
Using the ball bearing as the nucleus what would be the radius of the model?

My work/answer all values are in m(meters):
$1.5\times10^{-3} = x(1.2\times10^{-15})\\ x = \frac{1.5\times10^{-3}}{1.2\times10^{-15}} = 1.25\times10^{12}\\ r = x(0.529\times10^{-10}) = (1.25\times10^{12}) \times (0.529\times10^{-10}) = 66.125m$

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Done well. No change needed. I had automatically typed out a solution, but your is good, just post it and after a while, if nothing better comes up, accept it. –  André Nicolas Aug 29 '12 at 19:58
    
@AndréNicolas So you're saying my answer and work is correct? –  LF4 Aug 29 '12 at 20:22
    
Yes, sure. I wrote out essentially the same solution. So that you can see it, I will temporarily undelete it, then delete again. –  André Nicolas Aug 29 '12 at 20:31
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1 Answer 1

up vote 1 down vote accepted

Your calculation is fully correct. As per request, here is a slight variant, with details.

Let $r$ be the radius of the ball bearing, and let $R$ be the radius of the "atom" if the ball-bearing was the nucleus. Then $$\frac{R}{r}=\frac{0.529\times 10^{-10}}{1.2\times 10^{-15}}.$$ This expresses the fact we are looking at a scaled version of the atom/nucleus. But we know $r$, so we can find $R$. We get $$R=r\frac{0.529\times 10^{-10}}{1.2\times 10^{-15}}=(1.5\times 10^{-3})\frac{0.529\times 10^{-10}}{1.2\times 10^{-15}}.$$ For the calculation, I would first find $1.5\frac{0.529}{1.2}$, which is $0.66125$, and then deal with the power of $10$, which is $-3+(-10)-(-15)$, which is $2$. So the answer, in standard scientific notation, is $6.6125\times 10^1$ (metres), though $66.125$ seems more sensible. Many Physics people would correctly complain that we should only give the answer to $2$ significant figures, since two of the important numbers were only specified to that precision.

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