To determine the intensity level in decibels of a sound with an intensity of [tex]\(10^{-3}\)[/tex] watts per square meter (W/m²), given that the threshold of human hearing is [tex]\(1.0 \times 10^{-12}\)[/tex] W/m², we can use the following formula:
[tex]\[
L = 10 \log_{10} \left( \frac{I}{I_0} \right)
\][/tex]
where:
- [tex]\(L\)[/tex] is the intensity level in decibels (dB),
- [tex]\(I\)[/tex] is the intensity of the sound,
- [tex]\(I_0\)[/tex] is the reference intensity (threshold of human hearing).
Given values:
- [tex]\(I = 10^{-3} \, \text{W/m}^2\)[/tex]
- [tex]\(I_0 = 1.0 \times 10^{-12} \, \text{W/m}^2\)[/tex]
Now, we substitute these values into the formula:
[tex]\[
L = 10 \log_{10} \left( \frac{10^{-3}}{1.0 \times 10^{-12}} \right)
\][/tex]
First, we calculate the ratio inside the logarithm:
[tex]\[
\frac{10^{-3}}{1.0 \times 10^{-12}} = 10^{-3} \div 10^{-12} = 10^{-3} \times 10^{12} = 10^{9}
\][/tex]
Next, we take the logarithm base 10 of [tex]\(10^{9}\)[/tex]:
[tex]\[
\log_{10} (10^9) = 9
\][/tex]
Then we multiply by 10:
[tex]\[
L = 10 \times 9 = 90
\][/tex]
Therefore, the intensity level of the sound is:
[tex]\[
\boxed{90 \text{ dB}}
\][/tex]
So the correct answer is [tex]\(90 \text{ dB}\)[/tex].
