We want to look at photoelectric experiment in this problem. A simple vacuum tub

Question

We want to look at photoelectric experiment in this problem. A simple vacuum tube system for photoelectric experiment is shown, l he electrode under light illumination is made of Tungsten (W) with a work function of 4.5 eV. We use light sources with wavelengths 230 and 400nm to study photoelectric effect. What are frequency (Hz) and energy (eV) of photons for each light source? Calculate the voltage V required for reducing the photocurrent to zero as a function of the frequency of the light source.

Explanation / Answer

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 KEmax is the maximum kinetic energy of the emitted photoelectrons, and f is 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 energy hn are incident upon the cathode of a vacuum tube. The electrons in the cathode use a minimum f of 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. KEmax can 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 When solved for V, the equation becomes: V = (h/e) n - ( f /e) Thus, a plot of V versus ? for different frequencies of light will yield a linear plot with a slope (h/e) and a V intercept of (- f /e). may be this will help u

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