Question
Find the density of a hydrogen nucleus in grams per cubic centimeters, given that the diameter of this nucleus is 1.0 X 10-3pm and the mass is 1.67 X 10-24g. (HINT: Volume of a sphere of radius r is 4/3πr3.)
Answers
GPT-4o mini
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:
\[
\boxed{3.19 \times 10^{15} \, \text{g/cm}^3}
\]
\[
\text{Density} = \frac{\text{Mass}}{\text{Volume}}
\]
### 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:
\[
\boxed{3.19 \times 10^{15} \, \text{g/cm}^3}
\]