`""_(82)^(207)Pb` has a mass of `3.4368xx10^(-25)kg`. What is the approxiamte density of this lead nucleus?

Options

A. `2.3xx10^(17)kg//m^(3)`
B. `4.8xx10^(19)kg//m^(3)`
C. `3.5xx10^(18)kg//m^(3)`
D. `5.2xx10^(20)kg//m^(3)`

Answer

To find the approximate density of the lead nucleus given its mass, we can follow these steps:

### Step 1: Understand the formula for density
The density (ρ) of an object is defined as its mass (m) divided by its volume (V):
\[ \rho = \frac{m}{V} \]

### Step 2: Identify the mass of the nucleus
From the question, we know that the mass of the lead nucleus \( _{82}^{207}Pb \) is given as:
\[ m = 3.4368 \times 10^{-25} \text{ kg} \]

### Step 3: Calculate the volume of the nucleus
The volume of a nucleus can be approximated using the formula for the volume of a sphere:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the nucleus. The radius can be estimated using the empirical formula:
\[ r = r_0 A^{1/3} \]
where \( r_0 \) is a constant (approximately \( 1.2 \times 10^{-15} \) m) and \( A \) is the mass number (207 for lead).

### Step 4: Calculate the radius of the lead nucleus
Substituting the values:
\[ r = 1.2 \times 10^{-15} \times (207)^{1/3} \]
Calculating \( (207)^{1/3} \):
\[ (207)^{1/3} \approx 6.008 \]
Now calculate the radius:
\[ r \approx 1.2 \times 10^{-15} \times 6.008 \approx 7.21 \times 10^{-15} \text{ m} \]

### Step 5: Calculate the volume of the nucleus
Now substitute the radius back into the volume formula:
\[ V = \frac{4}{3} \pi (7.21 \times 10^{-15})^3 \]
Calculating \( (7.21 \times 10^{-15})^3 \):
\[ (7.21 \times 10^{-15})^3 \approx 3.74 \times 10^{-43} \text{ m}^3 \]
Now calculate the volume:
\[ V \approx \frac{4}{3} \times 3.14 \times 3.74 \times 10^{-43} \approx 1.57 \times 10^{-42} \text{ m}^3 \]

### Step 6: Calculate the density of the nucleus
Now we can find the density:
\[ \rho = \frac{m}{V} = \frac{3.4368 \times 10^{-25}}{1.57 \times 10^{-42}} \approx 2.19 \times 10^{17} \text{ kg/m}^3 \]

### Step 7: Conclusion
The approximate density of the lead nucleus is:
\[ \rho \approx 2.19 \times 10^{17} \text{ kg/m}^3 \]