Physics, asked by henals9302, 1 year ago

Why peaks (of same order) in X-ray diffraction have different intensity?

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Answered by KartikSharma13
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ANSWER--


OBJECT:

To study the emission of electrons from a metal surface which is irradiated with light. To experimentally determine the value of Planck's constant "h" by making use of the spectral dependency of the photoelectric effect.

APPARATUS:

A Mercury Vapor Light Source, a specialh/eapparatus with a photodiode tube, a digital voltmeter, a digital oscilloscope, green and yellow filters, and a New Focus variable transmission filter wheel.

BACKGROUND:

In 1901 a German physicist Max Planck published his law of radiation. Planck went on to state that the energy lost or gained by an oscillator is emitted or absorbed as a quantum of radiant energy, the magnitude of which is expressed by the equation:

E = hn

where E equals the radiant energy, n is the frequency of radiation, and h is a fundamental constant, now known as Planck's constant. Albert Einstein applied Planck's theory and explained the photoelectric effect in terms of the quantum model using his famous equation for which he received the Nobel Prize in 1921:

E = hn= KEmax+f

where KEmaxis the maximum kinetic energy of the emitted photoelectrons, andfis the energy needed to remove them from the surface of the material (the work function). Here E is the energy supplied by the quantum of light known as a photon.

In the h/e experiment, light photons with energyhnare incident upon the cathode of a vacuum tube. The electrons in the cathode use a minimumfof their energy to escape, leaving the surface with a maximum energy of KEmax.Normally the emitted electrons reach the anode of the tube, and can be measured as the photoelectric current. However, by applying a reverse potential V between the anode and cathode, the photoelectric current can be stopped.KEmaxcan then be determined by measuring the minimum reverse potential needed to bring the photoelectric current to zero. Thus, Einstein's relation becomes:

hn= Ve ++f

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