To find the density of a hydrogen nucleus, we can use the formula for density, which is defined as mass divided by volume:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
Step 1: Calculate the Volume
Given:
- The radius of the hydrogen nucleus is \( r = 1.0 \times 10^{-3} \) pm (picometers).
First, we need to convert this radius into centimeters: 1 picometer (pm) is equal to \( 1 \times 10^{-12} \) meters and \( 1 \) cm is \( 1 \times 10^{-2} \) meters, so we can convert picometers to centimeters:
\[ r = 1.0 \times 10^{-3} , \text{pm} = 1.0 \times 10^{-3} , \text{pm} \times \frac{1 \times 10^{-12} , \text{m}}{1 , \text{pm}} \times \frac{1 , \text{cm}}{1 \times 10^{-2} , \text{m}} = 1.0 \times 10^{-3} \times 10^{-12} , \text{cm} = 1.0 \times 10^{-15} , \text{cm} \]
Now, we can calculate the volume \( V \) of the nucleus, which is approximated as a sphere:
\[ V = \frac{4}{3} \pi r^3 \]
Substituting the value of \( r \):
\[ V = \frac{4}{3} \pi (1.0 \times 10^{-15} , \text{cm})^3 \]
Calculating \( r^3 \):
\[ (1.0 \times 10^{-15})^3 = 1.0 \times 10^{-45} , \text{cm}^3 \]
Now substituting back into the volume formula:
\[ V = \frac{4}{3} \pi (1.0 \times 10^{-45}) \approx \frac{4}{3} \times 3.14 \times 1.0 \times 10^{-45} \] \[ V \approx \frac{4.18879}{3} \times 10^{-45} \approx 4.18879 \times 10^{-45} , \text{cm}^3 \approx 4.19 \times 10^{-45} , \text{cm}^3 \]
Step 2: Calculate the Density
The mass of the hydrogen nucleus is given as \( 1.67 \times 10^{-24} , \text{g} \).
Now we can calculate the density:
\[ \text{Density} = \frac{1.67 \times 10^{-24} , \text{g}}{4.19 \times 10^{-45} , \text{cm}^3} \]
Calculating the density:
\[ \text{Density} \approx 3.98 \times 10^{20} , \text{g/cm}^3 \]
Final Answer:
The density of a hydrogen nucleus is approximately \( 3.98 \times 10^{20} , \text{g/cm}^3 \).