# 6.1 Blackbody radiation  (Page 6/15)

 Page 6 / 15

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?

about 3 K

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?

#### Questions & Answers

plot a graph of MP against tan ( Angle/2) and determine the slope of the graph and find the error in it.
Ime Reply
expression for photon as wave
BARISUA Reply
Are beta particle and eletron are same?
Amalesh Reply
yes
mari
how can you confirm?
Amalesh
sry
Saiaung
If they are same then why they named differently?
Amalesh
because beta particles give the information that the electron is ejected from the nucleus with very high energy
Absar
what is meant by Z in nuclear physic
Shubhu Reply
atomic n.o
Gyanendra
no of atoms present in nucleus
Sanjana
Note on spherical mirrors
Shamanth Reply
what is Draic equation? with explanation
M.D Reply
what is CHEMISTRY
trpathy Reply
it's a subject
Akhter
it's a branch in science which deals with the properties,uses and composition of matter
Eniabire
what is a Higgs Boson please?
FRANKLINE Reply
god particles is know as higgs boson, when two proton are reacted than a particles came out which is used to make a bond between than materials
M.D
bro little abit getting confuse if i am wrong than please clarify me
M.D
the law of refraction of direct current lines at the boundary between two conducting media of
BATTULA Reply
what is the black body of an ideal radiator
Areej Reply
uncertainty principles is applicable to
Areej
fermions
FRANKLINE
what is the cause of the expanding universe?
FRANKLINE
microscopic particles or gases
Areej
Astronomers theorize that the faster expansion rate is due to a mysterious, dark force that is pulling galaxies apart. One explanation for dark energy is that it is a property of space.
Areej
Thanks for your contribution Areej.
FRANKLINE
no problem
Areej
what is photoelectric equation
HIMANSHU Reply
How does fringe intensity depend upon slit width in single slit diffraction?
Abhishek Reply
intensity seems to be directly proportional radius of slit
Mathieu
what are the applications of Bernoulli's equation
Shaukat
VOLTE
Md Reply
what is Draic equation
M.D
about nuclear angular momentum
rahul Reply
what is spin
MUKESH Reply

### Read also:

#### Get the best University physics vol... course in your pocket!

Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'University physics volume 3' conversation and receive update notifications?

 By By Mistry Bhavesh