Scaling up the nucleus to the size of a pin.

[takando12]

Homework Statement

If the size of the nucleus ( in the range of 10^-15m to 10^-14m) is scaled up to the tip of a sharp pin, what roughly is the size of an atom ? Assume the tip of the pin to be in the range 10^-5m to 10^-4m

Homework Equations

The Attempt at a Solution

It's scaled up by a factor of 10¹⁰ i.e from 10^-15m to 10^-5m. And we also know that an atom's diameter is of the range 10^-10 m, and so 10^-10*10¹⁰ is 1 m. So if the nucleus is blown up to the size of a pin's tip, the size of the atom will be 1m right?
My textbook however says " Thus the nucleus of an atom is as small as the tip of a a sharp pin placed at the center of a sphere of radius 1 m. How is this ? The diameter of the sphere should be 1 m right? Where am i wrong in my logic?

 

[sjb-2812]

Not necessarily a fallacy. You are asked for the "order" of the size, not the "exact" size. What this means that as long as you have approximately the right figure you're OK and is a useful check. You may have made an error mathematically or otherwise and calculated this as a kilometre, or a milimetre, for instance and this is what the calculation is designed to catch

 

[takando12]

Not necessarily a fallacy. You are asked for the "order" of the size, not the "exact" size. What this means that as long as you have approximately the right figure you're OK and is a useful check. You may have made an error mathematically or otherwise and calculated this as a kilometre, or a milimetre, for instance and this is what the calculation is designed to catch

I don't understand. Is the radius of the sphere 1 m or not? I don't see where the calculation error is.please help.

 

[sjb-2812]

No error. It's approximately 1 metre, as opposed to 1 kilometre, or 1 millimetre.

 

[takando12]

No error. It's approximately 1 metre, as opposed to 1 kilometre, or 1 millimetre.

ok so if I hadn't seen the answer in my textbook and concluded that the radius of the sphere is 0.5 m , the answer is still fine right?

 

[sjb-2812]

I would think so, but just rereading the question, are you assuming that the figures given are radius, or diameter for both figures (minor point)

 

[takando12]

I would think so, but just rereading the question, are you assuming that the figures given are radius, or diameter for both figures (minor point)

Diameter is what I assumed. I also checked for the diameter of an atom and it's indeed in the order of 10^-10m and the diameter of the nucleus is 10^-15m as well. So do they mean radius when they say "size" in the question? They got it wrong?

 

[sjb-2812]

The "order" of something is an approximation. If given correct data you calculated the time it takes Usain Bolt to run 100m to be 0.1 seconds, 1 second, 10 seconds, 100 seconds, or 1000 seconds which is most likely to be correct?

But if you're taking both figures as diameter, where has the 0.5 come in?

 

[takando12]

The "order" of something is an approximation. If given correct data you calculated the time it takes Usain Bolt to run 100m to be 0.1 seconds, 1 second, 10 seconds, 100 seconds, or 1000 seconds which is most likely to be correct?

But if you're taking both figures as diameter, where has the 0.5 come in?

If we take both as diameters, then the 1 m I get after doing the multiplication must also be the diameter of the sphere with the pin in the middle right? And so in that case the radius must be 0.5m?

 

[Orodruin]

To state things mor bluntly: There is no one number which is a "correct" answer here. Atoms and nuclei come in many different sizes depending on the atomic number. As such you simply cannot give a precise value and bickering about 0.5 m vs 1 m is utterly pointless. Just as iff I ask you what the size of a boat is, you will not be able to answer 5.3 m. There are boats whicha are both much bigger and much smaller than this and thhe answer "of the order of 10 m" would be about as good as you could do.

 

[takando12]

So conclusion, I just say that it'll be in the order of 1 m. Period.

 

[Orodruin]

Yes, a factor of two is irrelevant when talking about orders of magnitude.

Edit: You might view order of magnitude computations as "what would be a good standard measure for x?"

 

[takando12]

Yes that makes sense, thank you all for the help.