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Question
Do all the electrons that absorb a photon come out as photoelectrons?
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Solution
The photoelectric effect is the emission of electrons (called photo-electrons when light strikes a surface. To escape from the surface, the electron must absorb enough energy from the incident radiation to overcome the attraction of positive ions in the material of the surface.
The photoelectric effect is based on the principle of conservation of energy.
1. Two conducting electrodes, the anode (Q) and cathode (P) are enclosed in an evacuated glass tube as shown on next page.
2. The battery or other source of potential difference creates an electric field in the direction from anode to cathode.
3. Light of a certain wavelength or frequency falling on the surface of the cathode causes a current in the external circuit called photoelectric current.
4. As the potential difference increases, photoelectric current also increases till saturation is reached.
5. When polarity of the battery is reversed (i.e., plate Q is at negative potential w.r.t. plate P) electrons start moving back towards the cathode.
6. At a particular negative potential of plate Q, no electron will reach the plate Q and the current will become zero. This negative potential is called stopping potential denoted by V0. Maximum kinetic energy of photoelectrons in terms of stopping potential will therefore be Kmax = (|V0|) eV
So we conclude that in the photoelectric effect, we can observe that most electrons get scattered into the metal by absorbing a photon.
Therefore, all the electrons that absorb a photon don't come out as photoelectron. Only a few come out of metal whose energy becomes greater than the work function of the metal.
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RELATED QUESTIONS
(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?
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?
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.]
The work function for the following metals is given:
Na: 2.75 eV; K: 2.30 eV; Mo: 4.17 eV; Ni: 5.15 eV
Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away?
Light of intensity 10−5 W m−2 falls on a sodium photo-cell of surface area 2 cm2. Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?
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?
Is it always true that for two sources of equal intensity, the number of photons emitted in a given time are equal?
Light of wavelength λ falls on a metal with work-function hc/λ0. Photoelectric effect will take place only if
When stopping potential is applied in an experiment on photoelectric effect, no photoelectric is observed. This means that
When the intensity of a light source in increased,
(a) the number of photons emitted by the source in unit time increases
(b) the total energy of the photons emitted per unit time increases
(c) more energetic photons are emitted
(d) faster photons are emitted
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
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)
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)
Find the maximum magnitude of the linear momentum of a photoelectron emitted when a wavelength of 400 nm falls on a metal with work function 2.5 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)
In an experiment on photoelectric effect, the stopping potential is measured for monochromatic light beams corresponding to different wavelengths. The data collected are as follows:-
Wavelength (nm): 350 400 450 500 550
Stopping potential (V): 1.45 1.00 0.66 0.38 0.16
Plot the stopping potential against inverse of wavelength (1/λ) on a graph paper and find (a) Planck's constant (b) the work function of the emitter and (c) the threshold wavelength.
(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 monochromatic beam is 1.2 × 1015 times per second. Find the maximum kinetic energy of the photoelectrons when this light falls on a metal surface whose work function is 2.0 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)
Two monochromatic beams A and B of equal intensity I, hit a screen. The number of photons hitting the screen by beam A is twice that by beam B. Then what inference can you make about their frequencies?
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?
