(a) Assuming the dimensions of the nucleus and atom shown in...

(a) Assuming the dimensions of the nucleus and atom shown in Figure $2.10,$ what fraction of the volume of the atom is taken up by the nucleus? (b) Using the mass of the proton from Table 2.1 and assuming its diameter is $1.0 \times 10^{-15} \mathrm{m},$ calculate the density of a proton in $\mathrm{g} / \mathrm{cm}^{3} .$

(a) Assuming the dimensions of the nucleus and atom shown in Figure $2.10,$ what fraction of the volume of the atom is taken up by the nucleus? (b) Using the mass of the proton
from Table 2.1 and assuming its diameter is $1.0 \times 10^{-15} \mathrm{m},$ calculate the density of a proton in $\mathrm{g} / \mathrm{cm}^{3} .$

Key Concepts

Atomic Structure

This concept involves understanding the structure of the atom, including the small, dense nucleus at its center and the much larger electron cloud surrounding it. In particular, the enormous difference in size between the nucleus and the entire atom is a fundamental feature of atomic structure that is critical in calculating volume fractions and density estimates.

Volume Calculation of Spherical Objects

Both the nucleus and the atom can be approximated as spheres. The calculation uses the volume formula for a sphere (V = 4/3?r³), which is a key mathematical tool in determining how much space a sphere occupies. This formula is essential for comparing the volume of the nucleus to that of the entire atom.

Unit Conversion

Unit conversion is crucial when working with measurements expressed in different unit systems. For example, converting lengths from meters to centimeters and masses from kilograms to grams ensures consistency, particularly when calculating densities in g/cm³. This concept is fundamental across scientific calculations to maintain precision and accuracy.

Density Calculation

Density is defined as mass per unit volume. In the context of this problem, the density of the proton is determined by dividing its mass by the volume derived from its diameter. This calculation illustrates the concept and importance of density in understanding how matter is distributed within a given space.

Transcript

00:01 All right, guys, we're doing problem 91 of chapter 2.

00:04 Assuming the dimensions of the atom and the nucleus shown in figure 2 .10, what fraction of the volume the atom is taken up by the nucleus? so we're going to use this, the volume of your radius, sorry, the volume of your nucleus over the volume of your atom is going to equal 4 3, 4 3 .4 .3 pi over the radius of your nucleus cube, and four -thirds pi times the radius of your atom cubed.

00:51 So you can see that these cancel out.

01:05 All right, guys, we're doing problem 91 of chapter 2.

01:08 Assuming the dimensions of the atom and the nucleus shown in figure 2 .10, what fraction of the volume the atom is taken up by the nucleus? so we're going to use this, the volume of your radius.

01:22 Sorry, the volume of your nucleus to the radius of your nucleus is, is going to be 10 to negative fourth divided by two, because 10 to the fourth's diameter of your nucleus cubed.

01:49 And they're both an angstrom, so i'm not going to bother writing the units.

01:54 Then your atom.

01:55 It can be one to five angstroms, so let's just take an average and put over the volume of your atom is going to equal 4 3, 4 thirds pi over the radius of your nucleus cubed, and 4 thirds pi times the radius of your atom cubed.

02:28 So you can see that these cancel out 3 divided by 2 cubed, and that'll equal 3 .7 times 10 to the negative.

02:59 14.

03:04 All right.

03:04 Next to the radius of your nucleus is is going to be 10 to negative fourth divided by two because 10 to the fourth's diameter of your nucleus cubed.

03:26 And they're both an angstrom, so i'm not going to bother writing the units.

03:30 Then your atom.

03:31 It can be one to five angstroms.

03:34 So let's just take an average and put three.

03:37 Next we are going to find the density of a proton.

03:42 And convert it to us as so we're going to find the density of a proton assuming its diameter is 1 .0 times 10 to 15 meters so we have their mass of proton convert from amu to grams that's going to give us 1 .67 10 to the negative 24 grams and then we divided by two cubed and that equal 3 .7 times 10 to the negative 14.

04:40 All right.

04:42 Next, we are going to find that we are going to convert this...

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