Wien's Law Explained

This question relates to determining the peak wavelength of radiation emitted by an object at a specific temperature, using Wien's Displacement Law. We need to compare the Sun's emission characteristics with those of a furnace.

Understanding Wien's Displacement Law

Wien's Displacement Law is a key concept in understanding thermal radiation. It states that the wavelength at which a black body emits the maximum intensity of radiation ($\lambda_{max}$) is inversely proportional to its absolute temperature ($T$). The formula is:

$$ \lambda_{max} T = b $$

Here:

• $\lambda_{max}$ represents the peak wavelength.
• $T$ is the absolute temperature in Kelvin (K).
• $b$ is Wien's displacement constant, approximately $2.898 \times 10^{-3}$ m·K.

It's important to use the temperature in Kelvin. The conversion from Celsius (°C) to Kelvin (K) is $ T (\text{K}) = T (°\text{C}) + 273.15 $. Also, note that the question uses units of μA (microamperes), which usually denotes current. However, in the context of radiation wavelength, the unit should be μm (micrometers). We will assume μA is a typo and proceed using μm for calculations.

Sun Radiation Analysis

We are given the Sun's surface temperature $ T_{sun} = 5800 \text{ K} $ and that it emits radiation at a wavelength $ \lambda_{sun} = 0.5 \text{ } \mu\text{m} $. Let's check if this is the peak wavelength using Wien's law:

$$ \lambda_{max, sun} = \frac{b}{T_{sun}} = \frac{2.898 \times 10^{-3} \text{ m·K}}{5800 \text{ K}} $$

$$ \lambda_{max, sun} \approx 5.00 \times 10^{-7} \text{ m} = 0.5 \text{ } \mu\text{m} $$

This confirms that 0.5 $\mu$m is the peak emission wavelength for the Sun under these conditions.

Furnace Wavelength Calculation

The furnace temperature is given as $ T_{furnace} = 300 \text{ } ^\circ\text{C} $. First, we must convert this temperature to Kelvin:

$$ T_{furnace} (\text{K}) = 300 + 273.15 = 573.15 \text{ K} $$

Now, we can use Wien's Displacement Law to find the peak wavelength ($ \lambda_{max, furnace} $) for the furnace:

$$ \lambda_{max, furnace} = \frac{b}{T_{furnace}} $$

Substituting the values:

$$ \lambda_{max, furnace} = \frac{2.898 \times 10^{-3} \text{ m·K}}{573.15 \text{ K}} $$

Performing the calculation:

$$ \lambda_{max, furnace} \approx 5.056 \times 10^{-6} \text{ m} $$

To match the units in the options, convert this result to micrometers:

$$ \lambda_{max, furnace} \approx 5.056 \text{ } \mu\text{m} $$

Result Comparison and Conclusion

The calculated peak radiation wavelength for the furnace is approximately 5.056 $\mu$m. We compare this result to the given options:

• 1. 5 $\mu$m
• 2. 0.5 $\mu$m
• 3. 2.5 $\mu$m
• 4. 0.25 $\mu$m

The value 5.056 $\mu$m is closest to 5 $\mu$m. Therefore, the furnace emits radiation at a peak wavelength of approximately 5 $\mu$m.