To determine the wavelength of the brightest color produced by a 3000 K object, we can use Wien's Displacement Law. The law is given by the formula:
[tex]\[ \lambda_{1\_max} \times T_1 = \lambda_{2\_max} \times T_2 \][/tex]
where:
- [tex]\(\lambda_{1\_max}\)[/tex] is the wavelength of the brightest color for the first object.
- [tex]\(T_1\)[/tex] is the temperature of the first object.
- [tex]\(\lambda_{2\_max}\)[/tex] is the wavelength of the brightest color for the second object.
- [tex]\(T_2\)[/tex] is the temperature of the second object.
Given values:
[tex]\[
T_1 = 6000 \, \text{K}
\][/tex]
[tex]\[
\lambda_{1\_max} = 483 \, \text{nm}
\][/tex]
[tex]\[
T_2 = 3000 \, \text{K}
\][/tex]
We need to find [tex]\(\lambda_{2\_max}\)[/tex].
First, rearrange the formula to solve for [tex]\(\lambda_{2\_max}\)[/tex]:
[tex]\[
\lambda_{2\_max} = \frac{\lambda_{1\_max} \times T_1}{T_2}
\][/tex]
Now, plug in the given values:
[tex]\[
\lambda_{2\_max} = \frac{483 \, \text{nm} \times 6000 \, \text{K}}{3000 \, \text{K}}
\][/tex]
Next, simplify the calculation:
[tex]\[
\lambda_{2\_max} = \frac{483 \times 6000}{3000}
\][/tex]
Finally, calculate the result:
[tex]\[
\lambda_{2\_max} = 966 \, \text{nm}
\][/tex]
Thus, the wavelength of the brightest color produced by a 3000 K object is [tex]\(966 \, \text{nm}\)[/tex]. The correct answer from the given options is:
[tex]\[
\boxed{966 \, \text{nm}}
\][/tex]
