Question

Consider a surface at \(500 \mathrm{~K}\). The spectral blackbody emissive power at a wavelength of \(50 \mu \mathrm{m}\) is (a) \(1.54 \mathrm{~W} / \mathrm{m}^{2} \cdot \mu \mathrm{m}\) (b) \(26.3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mu \mathrm{m}\) (c) \(108.4 \mathrm{~W} / \mathrm{m}^{2} \cdot \mu \mathrm{m}(d) 2750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mu \mathrm{m}(e) 8392 \mathrm{~W} / \mathrm{m}^{2} \cdot \mu \mathrm{m}\)

Short answer

Answer: The spectral blackbody emissive power at a wavelength of 50 µm for a surface at 500 K is approximately 1.54 W/m²·µm.

Step-by-step solution

Write down Planck's Law formula

Planck's Law describes the spectral distribution of blackbody radiation and can be written as: $$ E_\lambda(\lambda, T) = \frac{2 \pi h c^2}{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda k_BT}} - 1} $$ Where: \(E_\lambda\) is the spectral blackbody emissive power (\(\mathrm{W} / \mathrm{m}^{2} \cdot \mu \mathrm{m}\)), \(\lambda\) is the wavelength (\(\mathrm{m}\)), \(h\) is the Planck constant (\(6.626 \times 10^{-34} \mathrm{~Js}\)), \(c\) is the speed of light (\(2.998 \times 10^{8} \mathrm{~m/s}\)), \(T\) is the temperature in Kelvin (K), \(kB\) is the Boltzmann constant (\(1.381 \times 10^{-23} \mathrm{~J/K}\)).

Convert given wavelength to meters

The given wavelength is 50 µm, which we need to convert to meters for the calculation: $$ \lambda = 50 \times 10^{-6} \mathrm{~m} $$

Calculate the spectral blackbody emissive power

Now, we can plug in the given data into the Planck's Law formula to determine the spectral blackbody emissive power \(E_\lambda\): $$ E_\lambda = \frac{2 \pi (6.626 \times 10^{-34}) (2.998 \times 10^{8})^2}{(50 \times 10^{-6})^5} \frac{1}{e^{\frac{(6.626 \times 10^{-34})(2.998 \times 10^{8})}{(50 \times 10^{-6})(1.381 \times 10^{-23})(500)}} - 1} $$ After performing the calculations, we get: $$ E_\lambda \approx 1.54 \mathrm{~W} / \mathrm{m}^{2} \cdot \mu \mathrm{m} $$ The spectral blackbody emissive power at a wavelength of 50 µm for a surface at 500 K is approximately 1.54 \(\mathrm{W} / \mathrm{m}^{2} \cdot \mu \mathrm{m}\), which corresponds to option (a).

Key concepts

Spectral Blackbody Emissive Power

Spectral blackbody emissive power is a concept from thermal physics that explains how an idealized object emits radiation at different wavelengths. According to Planck's Law, every object emits thermal radiation differently based on its temperature, with a perfect blackbody being an ideal case. The power emitted per unit area per unit wavelength by a blackbody is calculated through Planck's Law, which considers factors such as wavelength and temperature. The formula is given by:\[E_\lambda(\lambda, T) = \frac{2 \pi h c^2}{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda k_BT}} - 1}\]Understanding this concept allows us to predict how much energy a blackbody will emit at a specific wavelength. This is crucial in fields like astrophysics and climate science, where such calculations help determine the properties of stars or the greenhouse effect of Earth's atmosphere.

Temperature in Kelvin

Temperature is a measure of heat energy within a substance. In thermodynamics, we often use the Kelvin scale, which is an absolute temperature scale based on theoretical concepts of absolute zero. Absolute zero is the point at which particles of matter stop moving entirely, representing a lack of thermal energy.
Kelvins provide an essential measure for scientific calculations because unlike Celsius or Fahrenheit, Kelvin is directly related to the thermodynamic laws. For example, 0 Kelvin represents absolute zero, thought to be unattainable in practice, but crucial for understanding energy distributions in systems.
In the context of blackbody radiation, Kelvin is standard for calculations because Planck's Law uses temperature in Kelvin to accurately determine spectral emissive power at given wavelengths. This allows for consistency across various scientific calculations and experiments.

Wavelength Conversion

Converting wavelengths into appropriate units is a key step in calculations involving Planck's Law. Wavelength is usually given in micrometers (µm) in terms of visible and infrared light, but it must be converted to meters for theoretical and mathematical processes.
To convert from micrometers to meters, simply multiply the wavelength by 10⁻⁶. For instance, a wavelength of 50 µm becomes 50 x 10⁻⁶ meters.
Understanding these conversions is crucial in accurate computation as it harmonizes units, preventing the errors that arise from unit mismatch. Proper conversions ensure that when substituting values into the formula for spectral blackbody emissive power, the calculations yield correct results, reflecting true physical phenomena.

Constants in Physics

In physics, constants are fundamental values that remain the same across various equations and experiments. They provide a reference to measure physical phenomena consistently and accurately.

• Planck's constant \(h\): A pivotal constant in quantum mechanics, \(h = 6.626 \times 10^{-34} \mathrm{~Js}\), describes the quantized nature of energy exchange.
• Speed of light \(c\): A fundamental constant representing the speed at which light travels in a vacuum, \(c = 2.998 \times 10^{8} \mathrm{~m/s}\).
• Boltzmann constant \(k_B\): Links temperature to energy, pivotal in statistical mechanics and thermodynamics, \(k_B = 1.381 \times 10^{-23} \mathrm{~J/K}\).

These constants are crucial in computations involving Planck's Law since they allow for the calculation of energy distributions from first principles, thereby providing deep insights into the behavior of blackbody radiation.