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प्रश्न
In the case of photoelectric effect experiment, explain the following facts, giving reasons.
The photoelectric current increases with increase of intensity of incident light.
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उत्तर
The intensity represents the number of photons, if the frequency of the incident light is more than the threshold frequency then more intensity will make ensure that more number of photons are falling over the metal surface and more photoelectrons will be emitted that increases the photoelectric current.
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संबंधित प्रश्न
Define the term 'intensity of radiation' in terms of photon picture of light.
A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used:
λ1 = 3650 Å, λ2 = 4047 Å, λ3 = 4358 Å, λ4 = 5461 Å, λ5 = 6907 Å,
The stopping voltages, respectively, were measured to be:
V01 = 1.28 V, V02 = 0.95 V, V03 = 0.74 V, V04 = 0.16 V, V05 = 0 V
Determine the value of Planck’s constant h, the threshold frequency and work function for the material.
[Note: You will notice that to get h from the data, you will need to know e (which you can take to be 1.6 × 10−19 C). Experiments of this kind on Na, Li, K, etc. were performed by Millikan, who, using his own value of e (from the oil-drop experiment) confirmed Einstein’s photoelectric equation and at the same time gave an independent estimate of the value of h.]
Draw graphs showing variation of photoelectric current with applied voltage for two incident radiations of equal frequency and different intensities. Mark the graph for the radiation of higher intensity.
An atom absorbs a photon of wavelength 500 nm and emits another photon of wavelength 700 nm. Find the net energy absorbed by the atom in the process.
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
Find the maximum kinetic energy of the photoelectrons ejected when light of wavelength 350 nm is incident on a cesium surface. Work function of cesium = 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)
The figure is the plot of stopping potential versus the frequency of the light used in an experiment on photoelectric effect. Find (a) the ratio h/e and (b) the work function.
Define the term: threshold frequency
In photoelectric effect, the photoelectric current started to flow. This means that the frequency of incident radiations is ______.
How would the stopping potential for a given photosensitive surface change if the frequency of the incident radiation were increased? Justify your answer.
The difference between threshold wavelengths for two metal surfaces A and B having work function ΦA = 9 eV and ΦB = 4.5 eV in nm is ______.
(Given, hc = 1242 eV nm)
