The maximum wavelength of emission from Neptune's cloud tops is 6.0 x 10-5m. Using this value, determine the temperature of Neptune's outer atmosphere.

Using Wein's displacement law, I get about 48K

lambdamax*temp=2.898E-3 m-Kelvin

Can you please show me how you calculated your answer?

Goodness. I used algebra I techniques.

Temp= 2.892E-3/6E-5= 48

To determine the temperature of Neptune's outer atmosphere using the maximum wavelength of emission from its cloud tops, we can make use of Wien's displacement law. This law states that the wavelength of peak emission (λ max) is inversely proportional to the temperature (T) of the object. The formula for Wien's displacement law is:

λ max = (b / T)

Where:
λ max is the maximum wavelength of emission
b is Wien's constant (2.898 × 10^-3 m·K)
T is the temperature in Kelvin (K)

Let's solve for T using the given maximum wavelength of emission:

λ max = 6.0 × 10^-5 m
b = 2.898 × 10^-3 m·K

Rearranging the formula, we have:

T = b / λ max

Substituting the given values:

T = (2.898 × 10^-3 m·K) / (6.0 × 10^-5 m)
T ≈ 48,300 K

Therefore, the approximate temperature of Neptune's outer atmosphere is 48,300 Kelvin.