# 6.1 Blackbody radiation  (Page 6/15)

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Check Your Understanding A molecule is vibrating at a frequency of $5.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{14}\text{Hz}.$ What is the smallest spacing between its vibrational energy levels?

$3.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-19}\text{J}$

## Quantum theory applied to a classical oscillator

A 1.0-kg mass oscillates at the end of a spring with a spring constant of 1000 N/m. The amplitude of these oscillations is 0.10 m. Use the concept of quantization to find the energy spacing for this classical oscillator. Is the energy quantization significant for macroscopic systems, such as this oscillator?

## Strategy

We use [link] as though the system were a quantum oscillator, but with the frequency f of the mass vibrating on a spring. To evaluate whether or not quantization has a significant effect, we compare the quantum energy spacing with the macroscopic total energy of this classical oscillator.

## Solution

For the spring constant, $k=1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\text{N/m},$ the frequency f of the mass, $m=1.0\text{kg},$ is

$f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}=\frac{1}{2\pi }\sqrt{\frac{1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\text{N/m}}{1.0\text{kg}}}\simeq 5.0\phantom{\rule{0.2em}{0ex}}\text{Hz}$

The energy quantum that corresponds to this frequency is

$\text{Δ}E=hf=\left(6.626\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-34}\text{J}·\text{s}\right)\left(5.0\text{Hz}\right)=3.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-33}\text{J}$

When vibrations have amplitude $A=0.10\text{m},$ the energy of oscillations is

$E=\frac{1}{2}k{A}^{2}=\frac{1}{2}\left(1000\text{N/m}\right){\left(0.1\text{m}\right)}^{2}=5.0\text{J}$

## Significance

Thus, for a classical oscillator, we have $\text{Δ}E\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}E\approx {10}^{-34}.$ We see that the separation of the energy levels is immeasurably small. Therefore, for all practical purposes, the energy of a classical oscillator takes on continuous values. This is why classical principles may be applied to macroscopic systems encountered in everyday life without loss of accuracy.

Check Your Understanding Would the result in [link] be different if the mass were not 1.0 kg but a tiny mass of 1.0 µ g, and the amplitude of vibrations were 0.10 µ m?

No, because then $\text{Δ}E\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}E\approx {10}^{-21}$

When Planck first published his result, the hypothesis of energy quanta was not taken seriously by the physics community because it did not follow from any established physics theory at that time. It was perceived, even by Planck himself, as a useful mathematical trick that led to a good theoretical “fit” to the experimental curve. This perception was changed in 1905 when Einstein published his explanation of the photoelectric effect, in which he gave Planck’s energy quantum a new meaning: that of a particle of light.

## Summary

• All bodies radiate energy. The amount of radiation a body emits depends on its temperature. The experimental Wien’s displacement law states that the hotter the body, the shorter the wavelength corresponding to the emission peak in the radiation curve. The experimental Stefan’s law states that the total power of radiation emitted across the entire spectrum of wavelengths at a given temperature is proportional to the fourth power of the Kelvin temperature of the radiating body.
• Absorption and emission of radiation are studied within the model of a blackbody. In the classical approach, the exchange of energy between radiation and cavity walls is continuous. The classical approach does not explain the blackbody radiation curve.
• To explain the blackbody radiation curve, Planck assumed that the exchange of energy between radiation and cavity walls takes place only in discrete quanta of energy. Planck’s hypothesis of energy quanta led to the theoretical Planck’s radiation law, which agrees with the experimental blackbody radiation curve; it also explains Wien’s and Stefan’s laws.

## Conceptual questions

Which surface has a higher temperature – the surface of a yellow star or that of a red star?

yellow

Describe what you would see when looking at a body whose temperature is increased from 1000 K to 1,000,000 K.

Explain the color changes in a hot body as its temperature is increased.

goes from red to violet through the rainbow of colors

Speculate as to why UV light causes sunburn, whereas visible light does not.

Two cavity radiators are constructed with walls made of different metals. At the same temperature, how would their radiation spectra differ?

would not differ

Discuss why some bodies appear black, other bodies appear red, and still other bodies appear white.

If everything radiates electromagnetic energy, why can we not see objects at room temperature in a dark room?

human eye does not see IR radiation

How much does the power radiated by a blackbody increase when its temperature (in K) is tripled?

## Problems

A 200-W heater emits a 1.5-µm radiation. (a) What value of the energy quantum does it emit? (b) Assuming that the specific heat of a 4.0-kg body is $0.83\text{kcal}\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}\text{kg}·\text{K},$ how many of these photons must be absorbed by the body to increase its temperature by 2 K? (c) How long does the heating process in (b) take, assuming that all radiation emitted by the heater gets absorbed by the body?

a. 0.81 eV; b. $2.1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{23};$ c. 2 min 20 sec

A 900-W microwave generator in an oven generates energy quanta of frequency 2560 MHz. (a) How many energy quanta does it emit per second? (b) How many energy quanta must be absorbed by a pasta dish placed in the radiation cavity to increase its temperature by 45.0 K? Assume that the dish has a mass of 0.5 kg and that its specific heat is $0.9\phantom{\rule{0.2em}{0ex}}\text{kcal}\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}\text{kg}·\text{K}.$ (c) Assume that all energy quanta emitted by the generator are absorbed by the pasta dish. How long must we wait until the dish in (b) is ready?

(a) For what temperature is the peak of blackbody radiation spectrum at 400 nm? (b) If the temperature of a blackbody is 800 K, at what wavelength does it radiate the most energy?

a. 7245 K; b. 3.62 μm

The tungsten elements of incandescent light bulbs operate at 3200 K. At what frequency does the filament radiate maximum energy?

Interstellar space is filled with radiation of wavelength $970\text{μ}\text{m.}$ This radiation is considered to be a remnant of the “big bang.” What is the corresponding blackbody temperature of this radiation?

The radiant energy from the sun reaches its maximum at a wavelength of about 500.0 nm. What is the approximate temperature of the sun’s surface?

how does colour appear in thin films
in the wave equation y=Asin(kx-wt+¢) what does k and w stand for.
derivation of lateral shieft
Hi
Hi
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ALFRED
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hi
asif
hi
Imran
I'm fine
ALFRED
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Anything which changes itself with respect to time or surrounding
Sushant
good
Chemist
and what's time? is time everywhere same
Chemist
No
Sushant
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do u know about black hole
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Not so more
Sushant
DHEERAJ
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But ask anything changes itself with respect to time or surrounding A Not any harmful radiation
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acceleration also inc
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acceleration also increases
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there are two correct answers depending on whether air resistance is considered. none of those answers have acceleration increasing.
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Acceleration is the change in velocity over time, hence it's the derivative of the velocity with respect to time. So this case would depend on the velocity. More specifically the change in velocity in the system.
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40 cm into change mm
40cm=40.0×10^-2m =400.0×10^-3m =400mm. that cap(^) I have used above is to the power.
Prema
i.e. 10to the power -2 in the first line and 10 to the power -3 in the the second line.
Prema
there is mistake in my first msg correction is 40cm=40.0×10^-2m =400.0×10^-3m =400mm. sorry for the mistake friends.
Prema
40cm=40.0×10^-2m =400.0×10^-3m =400mm.
Prema
this msg is out of mistake. sorry friends​.
Prema
what is physics?
why we have physics
because is the study of mater and natural world
John
because physics is nature. it explains the laws of nature. some laws already discovered. some laws yet to be discovered.
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physics is the study of non living things if we added it with biology it becomes biophysics and bio is the study of living things tell me please what is this?
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physics is the study of matter,energy and their interactions
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tahreem
thanks buvanas
tahreem
is this a physics forum