Which Isotope Has Half the Nuclear Radius of Dysprosium-160?

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SUMMARY

The discussion centers on identifying an isotope with a nuclear radius that is half that of Dysprosium-160, which has an atomic number of 66 and an atomic mass of 159.952202 u. The correct answer is Neon (Ne), with an atomic mass of 20 and atomic number 10. The radius of a nucleus is calculated using the formula r = (1.2 x 10^-15)(A^[1/3]), where A is the atomic mass. The initial calculations incorrectly used the atomic number instead of the atomic mass, leading to confusion in determining the unknown nucleus.

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  • Understanding of nuclear physics concepts, specifically nuclear radius calculations.
  • Familiarity with the formula for nuclear radius: r = (1.2 x 10^-15)(A^[1/3]).
  • Knowledge of atomic numbers and atomic masses of isotopes.
  • Basic algebra skills for manipulating equations.
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  • Explore additional examples of nuclear radius calculations using different isotopes.
  • Investigate the implications of nuclear radius on stability and decay of isotopes.
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This discussion is beneficial for students studying nuclear physics, educators teaching atomic structure, and anyone interested in the properties of isotopes and their calculations.

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Homework Statement



A particular isotope of dysprosium has atomic number 66 and atomic mass 159.952202 u. Identify the nucleus with a radius that is one half that of this isotope.

A. Ne (A=20, Z=10) <<<Correct Answer
B. O (A=16, Z=8)
C. Mn (A=50, Z=25)
D. Ni (A=60, Z=28)

Homework Equations


The only relevant equation I can think of is the equation of a nucleus's radius in terms of its atomic number:

r = (1.2X10^-15)(A^[1/3])

The Attempt at a Solution



My attempt has gone as follows:

1. Find the radius of the given dysprosium isotope by plugging it's atomic number into the given formula:

r1 = (1.2X10^-15)(66^[1/3])
r1 = 4.849488E-15 m

2. Now I find the radius of the unknown nucleus:

r2 = (1/2)r1
r2 = 2.424744E-15 m

3. Finally, by plugging the r2 value into the original equation, I should be able to solve for the atomic number of the unknown nucleus:

r2 = 2.424744E-15 m = (1.2X10^-15)(A2^[1/3])

A2 = 8.25

This is obviously not correct, it seems I'm missing a step; possibly one involving the atomic mass, since it is mentioned in the problem. Any help would be appreciated.

Thanks!
 
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r1 = (1.2X10^-15)(66^[1/3])
This is wrong. A is atomic mass, not atomic number.
 
rl.bhat said:
r1 = (1.2X10^-15)(66^[1/3])
This is wrong. A is atomic mass, not atomic number.

Ahhhhhh, of course. Thanks again rl.bhat!
 

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