Advertisements
Advertisements
Question
Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons?
Advertisements
Solution
The work function of a metal is the minimum energy required for a conduction electron to get out of the metal surface. All the electrons in an atom do not have the same energy level. When a ray having some photon energy is incident on a metal surface, the electrons come out from different levels with different energies. Hence, these emitted electrons show different energy distributions.
RELATED QUESTIONS
Define the term 'intensity of radiation' in terms of photon picture of light.
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?
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.
In an experiment on photoelectric effect, a photon is incident on an electron from one direction and the photoelectron is emitted almost in the opposite direction. Does this violate the principle of conservation of momentum?
It is found that yellow light does not eject photoelectrons from a metal. Is it advisable to try with orange light or with green light?
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
Calculate the number of photons emitted per second by a 10 W sodium vapour lamp. Assume that 60% of the consumed energy is converted into light. Wavelength of sodium light = 590 nm
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
A beam of white light is incident normally on a plane surface absorbing 70% of the light and reflecting the rest. If the incident beam carries 10 W of power, find the force exerted by it on the surface.
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
A totally reflecting, small plane mirror placed horizontally faces a parallel beam of light, as shown in the figure. The mass of the mirror is 20 g. Assume that there is no absorption in the lens and that 30% of the light emitted by the source goes through the lens. Find the power of the source needed to support the weight of the mirror.
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
Answer the following question.
Plot a graph of photocurrent versus anode potential for radiation of frequency ν and intensities I1 and I2 (I1 < I2).
On the basis of the graphs shown in the figure, answer the following questions :
(a) Which physical parameter is kept constant for the three curves?
(b) Which is the highest frequency among v1, v2, and v3?
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.
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?
Why it is the frequency and not the intensity of the light source that determines whether the emission of photoelectrons will occur or not? Explain.
How would the stopping potential for a given photosensitive surface change if the frequency of the incident radiation were increased? Justify your answer.
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 ______.
Plot a graph showing the variation of photoelectric current, as a function of anode potential for two light beams having the same frequency but different intensities I1 and I2 (I1 > I2). Mention its important features.
- Assertion (A): For the radiation of a frequency greater than the threshold frequency, the photoelectric current is proportional to the intensity of the radiation.
- Reason (R): Greater the number of energy quanta available, the greater the number of electrons absorbing the energy quanta and the greater the number of electrons coming out of the metal.
