To find the density of a hydrogen nucleus, we need to use the formula for density, which is mass divided by volume:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
Step 1: Calculate the radius from the diameter
The diameter of the hydrogen nucleus is given as \(1.0 \times 10^{-3} , \text{pm}\) (picometers). The radius \(r\) is half of the diameter:
\[ r = \frac{1.0 \times 10^{-3} , \text{pm}}{2} = 0.5 \times 10^{-3} , \text{pm} \]
Step 2: Convert picometers to centimeters
1 picometer (pm) is \(1 \times 10^{-12}\) meters, and since \(1 , \text{cm} = 10^{-2} , \text{m}\):
\[ 1 , \text{pm} = 1 \times 10^{-12} , \text{m} = 1 \times 10^{-10} , \text{cm} \]
Thus, \[ 0.5 \times 10^{-3} , \text{pm} = 0.5 \times 10^{-3} \times 10^{-10} , \text{cm} = 0.5 \times 10^{-13} , \text{cm} \]
Step 3: Calculate the volume of the nucleus using the sphere volume formula
The volume \(V\) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
Plugging in \(r = 0.5 \times 10^{-13} , \text{cm}\):
\[ V = \frac{4}{3} \pi (0.5 \times 10^{-13})^3 \]
Calculating \( (0.5 \times 10^{-13})^3 \):
\[ (0.5)^3 = 0.125 \]
\[ (10^{-13})^3 = 10^{-39} \]
Thus,
\[ (0.5 \times 10^{-13})^3 = 0.125 \times 10^{-39} , \text{cm}^3 = 1.25 \times 10^{-40} , \text{cm}^3 \]
Now substituting back into the volume formula:
\[ V = \frac{4}{3} \pi (1.25 \times 10^{-40}) \approx \frac{4}{3} \times 3.14159 \times 1.25 \times 10^{-40} \]
Calculating this gives:
\[ V \approx \frac{4.18879 \times 1.25 \times 10^{-40}}{3} \approx 5.236 \times 10^{-40} , \text{cm}^3 \]
Step 4: Calculate the density
Now we can use the mass of the hydrogen nucleus, which is \(1.67 \times 10^{-24} , \text{g}\):
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} = \frac{1.67 \times 10^{-24} , \text{g}}{5.236 \times 10^{-40} , \text{cm}^3} \]
Calculating the density:
\[ \text{Density} \approx 3.19 \times 10^{15} , \text{g/cm}^3 \]
Final Answer
The density of a hydrogen nucleus is approximately:
\[ \boxed{3.19 \times 10^{15} , \text{g/cm}^3} \]