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प्रश्न
Explain how does (i) photoelectric current and (ii) kinetic energy of the photoelectrons emitted in a photocell vary if the frequency of incident radiation is doubled, but keeping the intensity same?
Show the graphical variation in the above two cases.
उत्तर
- The increase in the frequency of incident radiation has no effect on photoelectric current. This is because of incident photon of increased energy cannot eject more than one electron from the metal surface.
- The kinetic energy of the photoelectron becomes more than the double of its original energy. As the work function of the metal is fixed, so incident photon of higher frequency and hence higher energy will impart more energy to the photoelectrons.
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संबंधित प्रश्न
The following graph shows the variation of photocurrent for a photosensitive metal :
(a) Identify the variable X on the horizontal axis.
(b) What does the point A on the horizontal axis represent?
(c) Draw this graph for three different values of frequencies of incident radiation v1, v2 and v3 (v1 > v2 > v3) for same intensity.
(d) Draw this graph for three different values of intensities of incident radiation I1, I2 and I3 (I1 > I2 > I3) having same frequency.
Planck's constant has the same dimensions as
The equation E = pc is valid
Photoelectric effect supports quantum nature of light because
(a) there is a minimum frequency below which no photoelectrons are emitted
(b) the maximum kinetic energy of photoelectrons depends only on the frequency of light and not on its intensity
(c) even when the metal surface is faintly illuminated the photoelectrons leave the surface immediately
(d) electric charge of the photoelectrons is quantised
A photon of energy hv is absorbed by a free electron of a metal with work-function hv − φ.
When the sun is directly overhead, the surface of the earth receives 1.4 × 103 W m−2 of sunlight. Assume that the light is monochromatic with average wavelength 500 nm and that no light is absorbed in between the sun and the earth's surface. The distance between the sun and the earth is 1.5 × 1011 m. (a) Calculate the number of photons falling per second on each square metre of earth's surface directly below the sun. (b) How many photons are there in each cubic metre near the earth's surface at any instant? (c) How many photons does the sun emit per second?
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
The electric field associated with a light wave is given by `E = E_0 sin [(1.57 xx 10^7 "m"^-1)(x - ct)]`. Find the stopping potential when this light is used in an experiment on photoelectric effect with the emitter having work function 1.9 eV.
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
Define the term: stopping potential in the photoelectric effect.
In photoelectric effect the photo current ______.
The figure shows a plot of stopping potential (V0) versus `1/lambda`, where λ is the wavelength of the radiation causing photoelectric emission from a surface. The slope of the line is equal to ______.
