Question

Calculate the peak wavelengths of a) the solar light received by Earth, and b) the light emitted by the Earth. Assume that the surface temperatures of the Sun and the Earth are \(5800 .\) K and \(300 . K\), respectively.

Short answer

Question: Calculate the peak wavelengths of the solar light received by Earth and the light emitted by the Earth if the temperature of the Sun is 5800 K and the temperature of Earth's surface is 300 K. Answer: The peak wavelength of the solar light received by Earth is \(5 \times 10^{-7}\,\text{m}\), and the peak wavelength of the light emitted by the Earth is \(9.66 \times 10^{-6}\,\text{m}\).

Step-by-step solution

Find the peak wavelength of the solar light received by Earth

To determine the peak wavelength of the solar light received by Earth, we use the temperature of the Sun (\(T_{sun} = 5800\,\text{K}\)). Using Wien's Displacement Law, we can calculate the peak wavelength as follows: $$\lambda_{sun} = \frac{b}{T_{sun}}$$

Calculate the peak wavelength of the solar light received by Earth

Now, we will substitute the values into the formula and calculate the peak wavelength of the solar light received by Earth: $$\lambda_{sun} = \frac{2.898 \times 10^{-3}\,\text{K}\cdot\text{m}}{5800\,\text{K}}$$ Solve for \(\lambda_{sun}\): $$\lambda_{sun} = 5 \times 10^{-7}\,\text{m}$$ So, the peak wavelength of the solar light received by Earth is \(5 \times 10^{-7}\,\text{m}\).

Find the peak wavelength of the light emitted by the Earth

To determine the peak wavelength of the light emitted by the Earth, we use the temperature of the Earth (\(T_{earth} = 300\,\text{K}\)). Using Wien's Displacement Law, we can calculate the peak wavelength as follows: $$\lambda_{earth} = \frac{b}{T_{earth}}$$

Calculate the peak wavelength of the light emitted by the Earth

Now, we will substitute the values into the formula and calculate the peak wavelength of the light emitted by the Earth: $$\lambda_{earth} = \frac{2.898 \times 10^{-3}\,\text{K}\cdot\text{m}}{300\,\text{K}}$$ Solve for \(\lambda_{earth}\): $$\lambda_{earth} = 9.66 \times 10^{-6}\,\text{m}$$ So, the peak wavelength of the light emitted by the Earth is \(9.66 \times 10^{-6}\,\text{m}\).