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

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Given that
Size of range the nucleus = 10^{-15} \text{ to } 10^{-14} m
Tip of the pin = 10^{-5} \text{ to } 10^{-4} m
Actual size of the nucleus = 10^{-10} m
Imaginary figure:
Step 1: Calculating the scaling factor of the nucleus
\text{ Scaling factor }=\frac{\text{ New size of the nucleus }}{\text{ Original size of nucleus }}
= \frac{10^{-5}}{10^{-15}}
= 10^{10}
Hence, scaling up by a factor of = 10^{10} m
Step 2: Finding the nucleus's approximate size?
\text{ approximate size } = \text{ scale factor * Actual size }
\text{ approximate size } = 10^{-10}* 10^{10}
\text{ approximate size } = 1
Hence, nucleus's approximate size 1 m