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# (II) $(a)$ What is the approximate radius of a $\frac{64}{29} \mathrm{Cu}$ nucleus? (b) Approximately what is the value of $A$ for a nucleus whose radius is $3.9 \times 10^{-15} \mathrm{m}$ ?

## a. 4.8$* 10^{-15} \mathrm{m}$b. 34

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##### Andy C.

University of Michigan - Ann Arbor

LB
##### Aspen F.

University of Sheffield

### Video Transcript

in this video, we are estimating the approximate size off a copper nucleus. So that is part A. So we've been given that we have Cooper 64 29 on the equation we need full of this question is that the approximate approximate radius of the nucleus oh, is approximately equal to a 1.2 times 10 to the minus 15 times by a to the third. What a is the atomic mass, which is this top number here. That's a so to do. Part of the question, all we're going to do is sub in 64 into this equation here and see what we get out as all so we have ah, approximately equal 1.2 times. Turn to the minus 15 times 64 to the third. So I'll just put that into my calculator 1.2 64 to the third. When we get that the radius, it's approximately 4.8 times. Turn to the minus 15 meters. So that is part A. And then part B wants us to find the value of a but we have a nucleus with a radius of 3.9 times 10 to the minus 15. So now we're just doing the backwards version of what we've done before. Last time we had a and didn't have are and now we have our but don't have a So if I so been the are we've been given, which is 3.9 times 10 to the minus 15 we know that that is approximately equal to 1.2 times 10 to the minus 15 times by some A to the third. So if I divide both sides of this equation by 1.2 times 10 to minus 15 I'll make eight of the third the subject. So 3.9 times 10 to the minus 15 divided 1.2 times 10 to the minus 15 is equal to 3.25 And then, if I q both sides of this equation, I will get our value for a on that gives us the A is equal to 34.3. So we will say that this is approximately 30 for and that's the question done

University of Sheffield
##### Andy C.

University of Michigan - Ann Arbor

LB
##### Aspen F.

University of Sheffield