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How much graphene is required to coat a 5 inch (diameter) sphere if the graphene layer must be 35 nanometers thick?

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The volume $V$ needed to cover the surface is equal to the spherical shell volume with $d_{\text{inner}}=5 \text{ inch}=0.127\text{ m}$ and $d_{\text{outer}}=d_{\text{inner}}+2\cdot 35 \cdot 10^{-9}\text{ m}$. Remember to use everything in SI-units.

$$V=\frac{4}{3}\pi\left[\left(\frac{d_{\text{outer}}}{2}\right)^3-\left(\frac{d_{\text{inner}}}{2}\right)^3\right]\approx1.7734\cdot 10^{-9} \text{ m}^3$$

If the density of graphene is given by $\rho$, then the necessary mass $m$ of graphene is given by $m=\rho V$.

Note, that the answer is very similar to the result obtained by Angus Murray, as the outer diameter is almost the same as the inner diameter. But I would not use approximative formulas as long as there is an exact formula because it will not be necessary to estimate the error that you have introduced by using this approximation.

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Surface area of the sphere is given by: $$A=\pi d^2$$

In $mm$, the diameter is equal to: $$d=5\times25.4=127 mm$$

So the surface area is equal to about 50671 $mm^2$

Multiply this by the thickness of the graphene layer (with a suitable conversion from $nm$ to $mm$) to get a volume of 0.0177 $mm^3$.

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  • $\begingroup$ You could argue this isn't really a good answer, as the surface area gets bigger when you move farther apart from the center. An integral would give the correct answer, not the approximation. $\endgroup$ May 23 '17 at 7:05
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    $\begingroup$ @Chemi: Seriously!? Yes, the integral is provides the mathematically correct method, but it really should be obvious from inspection that for a 35 nm layer on a 5 inch sphere, the fixed-radius approximation is quite valid. We do engineering here, and part of good engineering is to recognize when useful approximations can be applied. $\endgroup$ May 23 '17 at 11:36
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    $\begingroup$ The assumption that the graphene layer adds virtually no additional size to the sphere leads to an answer that is accurate to 6 significant figures - compare with the other two answers provided here. It is an assumption so obvious it is almost not worth mentioning $\endgroup$ May 24 '17 at 1:10
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$$ A = 4\pi r^2$$

$$ r = \frac{5 inch}{2} = \frac{127000000 nm}{2} = 63500000nm $$

$$ V = \int_{63500000}^{63500035} 4\pi r^2 $$

This turns out to be around 1.773 mm³.

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    $\begingroup$ You could argue this isn't really a good answer, as good engineering includes recognizing when simplifying approximations are valid and useful. Your doing the integral to account for different surface area at 5 inch radius and 5 inch plus 35 nm is absurd in this context. $\endgroup$ May 23 '17 at 11:39
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    $\begingroup$ Graphene is quite expensive a material. I would imagine if you were to coat thousands of balls a day, there would be a difference at the end of the month. Also, @MrYouMath 's answer does the integral too. It's better to be precise, imo. I was just giving another answer, no need to get jumpy. Have a nice day! $\endgroup$ May 23 '17 at 11:43
  • $\begingroup$ The reason I responded to this answer is because how you responded to another competing answer. Trying to make it look bad when the method yields the same result to several more digits than the values in the original question was inappropriate. $\endgroup$ May 23 '17 at 11:47
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    $\begingroup$ I didn't intend that at all. I was just giving some advice. And also, the final answer given is plainly wrong, but some orders of magnitude. However, disregarding that fact, I only wanted the OP to know that he may be missing some info. Sorry if I deeply offended you or if you misunderstood my intention. In fact, I had no account here, so the only reason I created the account on Engineering.SE was to explain why I though that answer was wrong and give an answer myself; I have no other interest. $\endgroup$ May 23 '17 at 11:56

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