Given properties of nucleus, find its atomic number, mass, and element symbol.

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SUMMARY

The discussion focuses on identifying an unknown nucleus with 71 neutrons and twice the volume of a nickel nucleus. The user applied the formula for nuclear radius, r = (1.2X10^-15)(A^(1/3)), and the volume formula, V = (4/3)(pi)(r)^3, to derive the atomic mass number. The correct atomic mass number was determined to be 118 after resolving an error in cubing the equation. The final expression for the radius incorporates both neutrons and protons, leading to the equation r = (1.2X10^-15)[(71 + Z)^(1/3)], where Z represents the number of protons.

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Homework Statement



An unknown nucleus contains 71 neutrons and has twice the volume of the nickel http://www.webassign.net/www32/symImages/f/1/3dca243bf11d1adbce4ca198d1973f.gif nucleus. Identify the unknown nucleus.

Homework Equations



r = (1.2X10^-15)(A^(1/3))
V= (4/3)(pi)(r)^3

The Attempt at a Solution



I plugged 59 into the radius equation and got "4.6715957X10^-15 m"
I then plugged that r value into the volume equation which yielded "4.270555X10^-43"

Since the unknown nucleus is twice the volume of the Ni nucleus, I doubled that volume (getting 8.54111078X10^-43), plugged it into the volume equation to solve for r and got "5.8858 X 10^-15".

Then I plugged that into the original radius equation to solve for A. I keep getting "14.71" which doesn't make any sense to me.

Thanks in advance!
 
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Ah i figured it out, nvm.

When I got to the --4.904836 = Cuberoot(A)-- I was supposed to cube the left side of the equation (yeilding 118 as the atomic mass number) but I was tripling it.

Duh!

:-)
 
An unknown nucleus contains 71 neutrons
So for the first nucleus r = (1.2X10^-15)[(71 + Z)^(1/3)] where Z is the number of protons.
Instead of finding radius you can wrote V = k*r = k*A
 

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