Calculate the density of flourine nucleus supposing that the shape of the nucleus is spherical and its radius is `5 xx 10^(-13)`. (Mass of `F`=`19` amu)

Options

A. `6.02xx10^(3) g " cm"^(-3)`
B. `5xx10^(4) g " cm"^(-3)`
C. `6.02xx10^(3) g " cm"^(-3)`
D. `5xx10^(13) g " cm"^(-3)`

Explanation

To calculate the density of the fluorine nucleus, we will follow these steps:

### Step 1: Understand the Given Data
- The radius of the fluorine nucleus (r) = \(5 \times 10^{-13}\) cm
- The mass of fluorine (F) = 19 amu

### Step 2: Convert Mass from amu to grams
To convert the mass from atomic mass units (amu) to grams, we use the conversion factor:
1 amu = \(1.66 \times 10^{-24}\) grams.

Thus, the mass of fluorine in grams is calculated as follows:
\[ \text{Mass} = 19 \, \text{amu} \times 1.66 \times 10^{-24} \, \text{g/amu} = 31.54 \times 10^{-24} \, \text{g} \]

### Step 3: Calculate the Volume of the Nucleus
The volume \(V\) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
Substituting the radius into the formula:
\[ V = \frac{4}{3} \pi (5 \times 10^{-13})^3 \]
Calculating \( (5 \times 10^{-13})^3 \):
\[ (5 \times 10^{-13})^3 = 125 \times 10^{-39} = 1.25 \times 10^{-37} \, \text{cm}^3 \]
Now substituting this back into the volume formula:
\[ V = \frac{4}{3} \times 3.14 \times 1.25 \times 10^{-37} \approx 5.24 \times 10^{-37} \, \text{cm}^3 \]

### Step 4: Calculate the Density
Density (\(D\)) is defined as mass per unit volume:
\[ D = \frac{\text{Mass}}{\text{Volume}} \]
Substituting the values we calculated:
\[ D = \frac{31.54 \times 10^{-24} \, \text{g}}{5.24 \times 10^{-37} \, \text{cm}^3} \]
Calculating this gives:
\[ D \approx 6.01 \times 10^{13} \, \text{g/cm}^3 \]

### Final Answer
The density of the fluorine nucleus is approximately \(6.01 \times 10^{13} \, \text{g/cm}^3\).

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