Advertisements
Advertisements
प्रश्न
Define the term 'intensity of radiation' in terms of photon picture of light.
Define intensity of radiation on the basis of photon picture of light.
उत्तर १
The intensity of radiation is defined as the energy associated with photons emitted from a unit surface area in unit time
उत्तर २
Intensity of radiation is defined as the number of photons falling per unit area in unit time.
APPEARS IN
संबंधित प्रश्न
(a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its e/m is given to be 1.76 × 1011 C kg−1.
(b) Use the same formula you employ in (a) to obtain electron speed for an collector potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified?
It is found that photosynthesis starts in certain plants when exposed to sunlight, but it does not start if the plants are exposed only to infrared light. Explain.
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
The work function of a photoelectric material is 4.0 eV. (a) What is the threshold wavelength? (b) Find the wavelength of light for which the stopping potential is 2.5 V.
(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.
In the case of photoelectric effect experiment, explain the following facts, giving reasons.
The photoelectric current increases with increase of intensity of incident light.
Consider a metal exposed to light of wavelength 600 nm. The maximum energy of the electron doubles when light of wavelength 400 nm is used. Find the work function in eV.
Consider a 20 W bulb emitting light of wavelength 5000 Å and shining on a metal surface kept at a distance 2 m. Assume that the metal surface has work function of 2 eV and that each atom on the metal surface can be treated as a circular disk of radius 1.5 Å.
- Estimate no. of photons emitted by the bulb per second. [Assume no other losses]
- Will there be photoelectric emission?
- How much time would be required by the atomic disk to receive energy equal to work function (2 eV)?
- How many photons would atomic disk receive within time duration calculated in (iii) above?
- Can you explain how photoelectric effect was observed instantaneously?
The work function for a metal surface is 4.14 eV. The threshold wavelength for this metal surface is ______.
Read the following paragraph and answer the questions.
| The figure shows the variation of photoelectric current measured in a photocell circuit as a function of the potential difference between the plates of the photocell when light beams A, B, C and D of different wavelengths are incident on the photocell. Examine the given figure and answer the following questions: |
- Which light beam has the highest frequency and why?
- Which light beam has the longest wavelength and why?
- Which light beam ejects photoelectrons with maximum momentum and why?
