Bilingual edition: English is preserved and Chinese follows each unit. Terminology uses the confirmed v20260916 glossary; automated semantic review remains traceable.
中英双语版：英文原文完整保留，中文紧随对应单元；术语采用 v20260916 确认表，自动语义审校结果可追溯。

A Nickel's Worth of Radiation

一镍的辐射量
 As an example of the kind of things you can model with the Planck radiation formula, consider the following questions:
  1. How much radiant energy comes from a nickel at room temperature per second?
    在室温下，镍每秒发出多少辐射能？
  2. How many photons per second leave the nickel?
    每秒有多少个光子离开镍？
  3. What volume of air would it take to have energy equal to 1 second of radiation from the nickel?
    要产生等于镍辐射1秒的辐射能，需要多少体积的空气？
  4. What volume of the room would have radiation energy equal to 1 second of radiation from the nickel?
    什么体积的房间中的辐射能量等于镍的1秒辐射能量？
    作为可以使用普朗克辐射公式建模的示例，考虑以下问题：

Measured properties of the nickel are diameter = 2.14 cm, thickness 0.2 cm, mass 5.1 grams. This gives a volume of 0.719 cm3 and a surface area of 8.54 cm 2. This gives a calculated density of 7.09 gm/cm3 compared to the density of 8.9 gm/cm3 for the pure element from the periodic table. So either the material is not pure nickel, being alloyed with some lighter element, or the measurement of thickness is off, which is a real possibility.

镍的测量属性为直径2.14厘米，厚度0.2厘米，质量5.1克。这给出体积0.719立方厘米和表面积8.54平方厘米。计算出的密度为7.09克/立方厘米，与元素周期表中纯镍的密度8.9克/立方厘米相比，说明材料可能不是纯镍，可能掺杂了某种较轻的元素，或者厚度的测量有误，这确实是一个可能的情况。

1. How much radiant energy comes from a nickel at room temperature per second?

1. 在室温下，镍每秒发出多少辐射能量？

The radiation from the nickel's surface can be calculated from the Stefan Boltzmann law. The room temperature will be taken to be 22°C = 295 K. Assuming an ideal radiator for this estimate, the radiated power is

镍表面的辐射可以用斯特凡-玻尔兹曼定律计算。取室温为22°C = 295 K。假设为理想辐射体，辐射功率为

P = σAT4 = (5.67 x 10-8 W/m2K4)x(8.54 x 10-4 m2)x(295 K)4 = 0.367 watts.

P = σAT⁴ = (5.67 × 10⁻⁸ W/m²·K⁴)x(8.54 × 10⁻⁴ m²)x(295 K)⁴ = 0.367 瓦。

So the radiated power from a nickel at room temperature is about 0.37 watts.

室温下镍辐射的功率约为0.37瓦。

2. How many photons per second leave the nickel?

2. 每秒有多少个光子从镍中逸出？

Since we know the energy, we can divide it by the average photon energy. We don't know a true average, but the wavelength of the peak of the blackbody radiation curve is a representative value which can be used as an estimate. This may be obtained from the Wien displacement law.

既然我们知道能量，我们可以将其除以平均光子能量。我们不知道真正的平均值，但黑体辐射曲线峰值的波长是一个代表值，可以用来作为估计。这可能来自维恩位移定律。

λpeak = 0.0029 m K/295 K = 9.83 x 10-6 m = 9830 nm, in the infrared.

λ peak = 0.0029 m·K/295 K = 9.83×10⁻⁶ m = 9830 nm，处于红外线范围。

The energy per photon at this peak can be obtained from the Planck relationship.

在这一峰值处，每个光子的能量可通过普朗克关系式获得。

Ephoton = hυ = hc/λ = 1240 eV nm/ 9830 nm = 0.126 eV

光子的能量 E_photon = hυ = hc/λ = 1240 eV nm / 9830 nm = 0.126 eV

Then the number of photons per second is

然后每秒的光子数是

N = (0.367 J)/(0.126 eV x 1.6 x 10-19 J/eV) = 1.82 x 1019 photons

N = (0.367 J)/(0.126 eV × 1.6 × 10⁻¹⁹ J/eV) = 1.82 × 10¹⁹ photons

3. What volume of air would it take to have energy equal to 1 second of radiation from the nickel?

3. 要产生等于镍辐射1秒的辐射能量，需要多少体积的空气？

The translational kinetic energy of an air molecule can be obtained from equipartition of energy and the definition of the kinetic temperature.

空气分子的平动动能可通过能量均分定理和动能温度的定义获得。
Kinetic energyaverage
平均
= 3/2 kT = 3/2 (8.617 x 10-5 eV/K)(295 K) = 0.038 eV

动能 = 3/2 kT = 3/2 (8.617 x 10 eV/K)(295 K) = 0.038 eV

The number of molecules to equal one second's worth of radiation is then

要使分子数等于一秒钟的辐射量则为
Nmolecules
分子
= (0.367 J)/(0.038 eV x 1.6 x 10-19 J/eV) = 6.0 x 1019 molecules

N = (0.367 J)/(0.038 eV x 1.6 x 10 J/eV) = 6.0 x 10 molecules（专名或术语）

The volume per air molecule can be found by using the fact that a mole of an ideal gas occupies 22.4 liters at STP. Using the ideal gas law

利用一摩尔理想气体在标准温度和压力下占据22.4升这一事实，可以求得每个空气分子的体积。使用理想气体定律

Vair molecule = (22.4 x 10-3 m3)(295K/273K)/(6.02 x 1023 molecules/mole) = 4.02 x 10-26m3

空气分子的体积为 (22.4 × 10⁻³ m³)(295K/273K)/(6.02 × 10²³ molecules/mole) = 4.02 × 10⁻²⁶ m³

Volume of air to have 0.367 J =(4.02 x 10-26m3)(6.0 x 1019 molecules) = 2.41x 10-6m3

体积为0.367 J的空气体积为(4.02×10⁻²⁶ m³)(6.0×10¹⁹ molecules) = 2.41×10⁻⁶ m³

So the small volume of about 2.4 cm3 of air will have kinetic energy equal to the amount of energy radiated from the nickel per second.

因此，体积约为2.4 cm³的空气将具有等于镍每秒辐射能量的动能。

4. What volume of the room would have radiation energy equal to 1 second of radiation from the nickel?

4. 多少体积的房间中的辐射能量等于镍的1秒辐射能量？

The Stefan-Boltzmann relationship is also related to the energy density in the radiation in a given volume of space. The energy density is

斯特藩-玻尔兹曼关系也与给定体积空间中的辐射的能量密度有关。能量密度是

4σAT4/c = 4(5.67 x 10-8 W/m2K4)x(295 K)4/(3 x 108m/s)

4σAT/c = 4(5.67 x 10 -8 W/m²K⁴)x(295 K)⁴/(3 x 10⁸ m/s)

Energy density in radiation = 5.73 x 10-6J/m3

辐射的能量密度 = 5.73 x 10 -6 J/m 3

So the volume required is (0.367 J)/(5.73 x 10-6J/m3) = 64000 m3. This is the volume of a cube with dimension 40 meters per side.

所需体积为 (0.367 J)/(5.73 × 10⁻⁶ J/m³) = 64000 m³。这相当于一个边长为40米的立方体的体积。

The air molecules in a volume of air about three times the volume of the nickel will have kinetic energy equal to the radiation from the nickel, but it takes the radiation content of a large auditorium to equal that amount. The ratio of the energy per unit volume in the molecular kinetic energy is more than 1010 times that in radiation.

体积约为镍的三倍的空气中的空气分子的动能等于镍的辐射能，但要达到这种能量水平，需要整个大型礼堂的辐射能量。分子动能的单位体积能量比辐射的能量高10^10倍。
Applications of the Planck radiation formula
普朗克辐射公式应用
What about the division of energy between mass and radiation in the universe?
宇宙中能量如何分配在质量和辐射之间？
 Index

Blackbody radiation concepts

Heat transfer concepts
索引 黑体辐射概念 热传递概念
 
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