Find the order of magnitude of the number of atom in `1cm^(3)` of a solid. Given that the diameter of an atom is about `10^(-10)m`

Explanation

To find the order of magnitude of the number of atoms in \(1 \, \text{cm}^3\) of a solid, we can follow these steps:

### Step 1: Convert the volume from cm³ to m³
Given:
- Volume of the solid = \(1 \, \text{cm}^3\)

To convert cm³ to m³:
\[ 1 \, \text{cm}^3 = 10^{-6} \, \text{m}^3 \]

### Step 2: Calculate the volume of a single atom
Given:
- Diameter of an atom \(d = 10^{-10} \, \text{m}\)

The volume \(V\) of a single atom can be approximated as a sphere:
\[ V = \frac{4}{3} \pi r^3 \]
where \(r = \frac{d}{2} = \frac{10^{-10}}{2} = 5 \times 10^{-11} \, \text{m}\).

Now, substituting the radius into the volume formula:
\[ V = \frac{4}{3} \pi (5 \times 10^{-11})^3 \]
Calculating \( (5 \times 10^{-11})^3 \):
\[ (5 \times 10^{-11})^3 = 125 \times 10^{-33} = 1.25 \times 10^{-31} \, \text{m}^3 \]
Now, substituting this back into the volume formula:
\[ V \approx \frac{4}{3} \times 3.14 \times 1.25 \times 10^{-31} \]
Calculating this gives:
\[ V \approx 5.24 \times 10^{-31} \, \text{m}^3 \]

### Step 3: Calculate the number of atoms in the given volume
Now we can find the number of atoms \(N\) in \(1 \, \text{cm}^3\) of solid:
\[ N = \frac{\text{Volume of solid}}{\text{Volume of one atom}} = \frac{10^{-6} \, \text{m}^3}{5.24 \times 10^{-31} \, \text{m}^3} \]
Calculating this gives:
\[ N \approx 1.91 \times 10^{24} \]

### Step 4: Find the order of magnitude
The order of magnitude is determined by the exponent of 10 in the number:
\[ N \approx 1.91 \times 10^{24} \implies \text{Order of magnitude} = 24 \]

### Final Answer:
The order of magnitude of the number of atoms in \(1 \, \text{cm}^3\) of a solid is \(24\).

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