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E = n h ν , n = 1 , 2 , 3 , . . .

The quantity h is a constant now known as Planck's constant, in his honor. Although Planck was pleased he had resolved the blackbody radiation paradox, he was disturbed that to do so, he needed to assume the vibrating atoms required quantized energies, which he was unable to explain. The value of Planck's constant is very small, 6.626 × 10 −34 joule seconds (J s), which helps explain why energy quantization had not been observed previously in macroscopic phenomena.

A graph is shown with a horizontal axis labeled, “Wavelength lambda (micrometers)” and a vertical axis labeled, “Intensity I (a r b. units).” The horizontal axis begins at 0 and extends to 3.0 with markings provided every 0.1 micrometer. Similarly, the vertical axis begins at 0 and extends to 10 with markings every 1 unit. Two vertical dashed lines are drawn. The first appears at about 0.39 micrometers and the second at about 0.72 micrometers. To the left of the first of these lines, the label, “Ultraviolet,” appears at the top of the graph. Between these lines, the label, “Visible,” appears at the top of the graph. To the right of the second of these lines, the label, “Infrared,” appears at the top of the graph. To the far right of the graph in open space a purple dot is placed which is labeled, “lambda maximum.” A “Temperature” label is located in a central region of the graph. A blue curve begins on the horizontal axis at about 0.05 micrometers. This curve increases steeply to a maximum value between the two vertical line segments of approximately 9.5 at about 0.55 micrometers. This curve decreases rapidly at first, then tapers off to reach a value of about 1.5 at the far right end of the graph. This blue curve is labeled 6000 K beneath the “Temperature” label. Curves are similarly drawn in green for 5000 K, orange for 4000 K, and red for 3000 K. As the temperature decreases, the height of the peak is lower and shifted right on the graph. The maximum value for the green curve is around 4.5 at 7.2 micrometers. This curve tapers at the right end of the graph to a value around 0.6. The maximum for the orange curve is around 2 at about 0.9 micrometers. This curve tapers at the right end of the graph to a value around 0.2. The maximum for the red curve is around 0.7 at about 1.2 micrometers. This curve tapers at the right end of the graph to a value around 0.1. The entire region under the blue curve that is between the two dashed lines, indicating the visible region, is shaded with vertical bands of color. The colors extending left to right across this region are violet, indigo, blue, green, yellow, orange, and red. A purple dot is placed at the peak of each of the four colored curves. These peaks are connected with a dashed curve.
Blackbody spectral distribution curves are shown for some representative temperatures.

The photoelectric effect

The next paradox in the classical theory to be resolved concerned the photoelectric effect ( [link] ). It had been observed that electrons could be ejected from the clean surface of a metal when light having a frequency greater than some threshold frequency was shone on it. Surprisingly, the kinetic energy of the ejected electrons did not depend on the brightness of the light, but increased with increasing frequency of the light. Since the electrons in the metal had a certain amount of binding energy keeping them there, the incident light needed to have more energy to free the electrons. According to classical wave theory, a wave's energy depends on its intensity (which depends on its amplitude), not its frequency. One part of these observations was that the number of electrons ejected within in a given time period was seen to increase as the brightness increased. In 1905, Albert Einstein was able to resolve the paradox by incorporating Planck's quantization findings into the discredited particle view of light (Einstein actually won his Nobel prize for this work, and not for his theories of relativity for which he is most famous).

Einstein argued that the quantized energies that Planck had postulated in his treatment of blackbody radiation could be applied to the light in the photoelectric effect so that the light striking the metal surface should not be viewed as a wave, but instead as a stream of particles (later called photons ) whose energy depended on their frequency, according to Planck's formula, E = (or, in terms of wavelength using c = νλ , E = h c λ ). Electrons were ejected when hit by photons having sufficient energy (a frequency greater than the threshold). The greater the frequency, the greater the kinetic energy imparted to the escaping electrons by the collisions. Einstein also argued that the light intensity did not depend on the amplitude of the incoming wave, but instead corresponded to the number of photons striking the surface within a given time period. This explains why the number of ejected electrons increased with increasing brightness, since the greater the number of incoming photons, the greater the likelihood that they would collide with some of the electrons.

With Einstein's findings, the nature of light took on a new air of mystery. Although many light phenomena could be explained either in terms of waves or particles, certain phenomena, such as the interference patterns obtained when light passed through a double slit, were completely contrary to a particle view of light, while other phenomena, such as the photoelectric effect, were completely contrary to a wave view of light. Somehow, at a deep fundamental level still not fully understood, light is both wavelike and particle-like. This is known as wave-particle duality    .

Questions & Answers

what is the meaning of intermolecular force
Eunice Reply
is the force of attraction that exist between two or more molecules
Johnson
What is a primary standard solution ?
Duval
a known solution
Fiko
Characteristic of a primary standard solution
Duval
pauli's exclusion is based on what?
avdhesh Reply
What is greatest modification made in dalton's atomic theory?
Ngwesse Reply
Types of electrolytes
Treasure Reply
Strong, weak and non-electrolytes
Grace
welcome
Alieu
thanks what's this platform all about
Nnamdi
list 6 subatomic particles and their mass, speed and charges
Dubem Reply
combination of acid and base
Ayibiro Reply
that salt
Talhatu
calculate the mass in gram of NaOH present in 250cm3 of 0.1mol/dm3 of its solution
Omego Reply
The mass is 1.0grams. First you multiply the molecular weight and molarity which is 39.997g/mol x 0.1mol/dm3= 3.9997g/dm3. Then you convert dm3 to cm3. 1dm3 =1000cm3. In this case you would divide 3.9997 by 1000 which would give you 3.9997*10^-3 g/cm3. To get the mass you multiply 3.9997*10^-3 and
Kokana
250cm3 and get the mass as .999925, with significant figures the answer is 1.0 grams
Kokana
nitrogen, phosphorus, arsenic, antimony and Bismuth
faith Reply
What is d electronic configuration of for group 5
Miracle Reply
Can I know d electronic configuration of for group 5 elements
Miracle
2:5, 2:8:5, 2:8:8:5,...
Maxime
Thanks
Miracle
Pls what are d names of elements found in group 5
Miracle
define define. define
Muh Reply
what is enthalpy
Ayilaran Reply
total heat contents of the system is called enthalpy, it is state function.
Sajid
background of chemistry
Banji Reply
what is the hybridisation of carbon in formic acid?
Maham Reply
sp2 hybridization
Johnson
what is the first element
Josh Reply
HYDROGEN
Liklai
Element that has positive charge and its non metal Name the element
Liklai
helium
oga
sulphur
oga
hydrogen
Banji
account for the properties of organic compounds
mercy Reply
properties of organic compounds
mercy

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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