# 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?

is this a physics forum
explain l-s coupling
how can we say dirac equation is also called a relativistic equation in one word
what is the electronic configration of Al
what's the signeficance of dirac equetion.?
what is the effect of heat on refractive index
As refractive index depend on other factors also but if we supply heat on any system or media its refractive index decrease. i.e. it is inversely proportional to the heat.
ganesh
you are correct
Priyojit
law of multiple
Wahid
if we heated the ice then the refractive index be change from natural water
Nepal
can someone explain normalization condition
Swati
yes
Chemist
1 millimeter is How many metres
1millimeter =0.001metre
Gitanjali
The photoelectric effect is the emission of electrons when light shines on a material.
What is photoelectric effect
it gives practical evidence of particke nature of light.
Omsai
particle nature
Omsai
photoelectric effect is the phenomenon of emission of electrons from a material(i.e Metal) when it is exposed to sunlight. Emitted electrons are called as photo electrons.
Anil
what are the applications of quantum mechanics to medicine?
Neptune
application of quantum mechanics in medicine: 1) improved disease screening and treatment ; using a relatively new method known as BIO- BARCODE ASSAY we can detect disease-specific clues in our blood using gold nanoparticles. 2) in Genomic medicine 3) in protein folding 4) in radio theraphy(MRI)
Anil
Quantam physics ki basic concepts?
why does not electron exits in nucleaus
electrons have negative
YASH
Proton and meltdown has greater mass than electron. So it naturally electron will move around nucleus such as gases surrounded earth
Amalesh
.......proton and neutron....
Amalesh
excuse me yash what negative
Rika
coz, electron contained minus ion
Manish
negative sign rika shrestha ji
YASH
electron is the smallest negetive charge...An anaion i.e., negetive ion contains extra electrons. How ever an atom is neutral so it must contains proton and electron
Amalesh
yes yash ji
Rika
yes friends
Prema
koantam theory
Laxmikanta
yes prema
Rika
quantum theory tells us that both light and matter consists of tiny particles which have wave like propertise associated with them.
Prema
proton and nutron nuclear power is best than proton and electron kulamb force
Laxmikanta
what is de-broglie wave length?
Ramsuphal
plot a graph of MP against tan ( Angle/2) and determine the slope of the graph and find the error in it.
expression for photon as wave
Are beta particle and eletron are same?
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
beta particle is of two kind beta plus and beta minus ,beta minus is electron and beta plus is positron
Nepal
beta particles are none but positive charged electron
Priyojit
which is ejected from nucleus
Priyojit
when nutron converts to proton it splits to two part one proton and the minus part ejected as beta
Priyojit
both are same things only the difference is that beta partical ejected from the neuclus and electrons revolves round the neuclus.
Sibghat