What is the peak wavelength of light coming from a star with a temperature of 5300 K ? (Submit your answer in nanometers. Remember \( 1 \mathrm{~nm}=10^{-9} \mathrm{~m} \) )

To find the peak wavelength of light emitted by a star with a temperature of 5300 K, we can use Wien's displacement law, which states that the peak wavelength (\( \lambda_{\text{max}} \)) is inversely proportional to the temperature (T) of the black body radiation. The formula is given by: \[ \lambda_{\text{max}} = \frac{b}{T} \] where \( b \) (Wien's displacement constant) is approximately \( 2898 \, \mu m \cdot K \). Using this formula: \[ \lambda_{\text{max}} = \frac{2898 \, \mu m \cdot K}{5300 \, K} = \frac{2898 \times 10^{-6} \, m \cdot K}{5300 \, K} \approx 0.000547 \, m = 547 \, nm \] So, the peak wavelength of light coming from a star with a temperature of 5300 K is approximately 547 nm. Real-world application: Understanding the peak wavelength can help astronomers classify stars and their temperatures. A star like our Sun, which has a similar temperature, emits most of its energy in the visible spectrum, allowing it to be observed with the naked eye! So, when you gaze at a starry night, you can appreciate that the light reaching you is a cosmic portrait of different temperatures in space. Problem-solving tips: When using Wien’s law, ensure your temperature is in Kelvin and your answer is converted from meters to nanometers correctly by multiplying by \(10^9\). A common mistake is to confuse Kelvin with Celsius, which could lead to significant errors in calculated wavelengths!