Advertisements
Advertisements
प्रश्न
Define the terms "stopping potential' and 'threshold frequency' in relation to the photoelectric effect. How does one determine these physical quantities using Einstein's equation?
उत्तर
Stopping potential:
For a particular frequency of incident radiation, the minimum negative (retarding) potential V0 given to the anode plate for which the photocurrent stops or becomes zero is called the cut-off or stopping potential.
Threshold frequency:
There exists a certain minimum cut-off frequency ν0, for which the stopping potential is zero and below ν0 the electron emission is not possible.
This cut-off frequency is known as threshold frequency ν0, which is different for different metal. In the photoelectric effect, an electron absorbs a quantum of energy (hν ) of radiation. If this quantum of energy absorbed by electron exceeds the minimum energy required to come out of the metal surface by electron, the kinetic energy of the emitted electron is
`"K" = "hv" - phi` ...(1)
Where `phi` is the minimum energy for electron to come out of the metal, and is different for different electrons in the metal. The maximum kinetic energy of photoelectrons is given by
`"K""max" = "hv" - phi0` ...(2)
Where, `phi0 - ` work function or least value of φ equation (2) is known as Einstein's photoelectric equation.
Explanation of photoelectric effect with the help of Einstein's photoelectric equation
(i) According to equation (2), Kmax depends linearly on ν, and is independent of the intensity of radiation. This happens because, here, the photoelectric effect arises from the absorption of a single quantum of radiation by a single electron. The intensity of the radiation (that is proportional to the number of energy quanta per unit area per unit time) is irrelevant to this basic process.
(ii) Since Kmax must be non-negative, equation (2) implies that photoelectric emission is possible only if h ν > `phi0`.
or v > v0, where v0 = `"V"_0 = phi_0/"h"`
Thus, there exists a threshold frequency v0 `"V"_0 = phi_0/"h"` exists, below which photoelectric emission is not possible, and is independent of intensity.
(iii) As the intensity of radiation is proportional to the number of energy quanta per unit area per unit time. The greater the number of energy quanta available, the greater is the number of electrons absorbing the energy quanta, and therefore, the number of electrons coming out of the metal (for ν > ν0) is more and so is photoelectric current.
APPEARS IN
संबंधित प्रश्न
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.]
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?
Should the energy of a photon be called its kinetic energy or its internal energy?
Two photons of
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)
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)
A sphere of radius 1.00 cm is placed in the path of a parallel beam of light of large aperture. The intensity of the light is 0.5 W cm−2. If the sphere completely absorbs the radiation falling on it, find the force exerted by the light beam on the sphere.
(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 sphere of radius 1.00 cm is placed in the path of a parallel beam of light of large aperture. The intensity of the light is 0.5 W cm−2. If the sphere completely absorbs the radiation falling on it, Show that the force on the sphere due to the light falling on it is the same even if the sphere is not perfectly absorbing.
Do all the electrons that absorb a photon come out as photoelectrons?
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.