Question
Find the density of a hydrogen nucleus in grams per cubic centimeters, given that the decimal of this nucleus is 1.0×10^-3pm and the mass is 1.67×10^-24g (HINT: volume of a sphere of radius r is 4/3π^3
Answers
GPT-4o mini
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 \( 3.98 \times 10^{20} \, \text{g/cm}^3 \).
\[
\text{Density} = \frac{\text{Mass}}{\text{Volume}}
\]
### 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 \( 3.98 \times 10^{20} \, \text{g/cm}^3 \).