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प्रश्न
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?
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उत्तर
Mo and Ni will not show photoelectric emission in both cases
Wavelength for a radiation, λ = 3300 Å = 3300 × 10−10 m
Speed of light, c = 3 × 108 m/s
Planck’s constant, h = 6.6 × 10−34 Js
The energy of incident radiation is given as:
`"E" = "hc"/lambda`
= `(6.6 xx 10^(-34) xx 3 xx 10^8)/(3300 xx 10^(-10))`
= 6 × 10−19 J
= `(6 xx 10^(-19))/(1.6 xx 10^(-19))`
= 3.158 eV
It can be observed that the energy of the incident radiation is greater than the work function of Na and K only. It is less for Mo and Ni. Hence, Mo and Ni will not show photoelectric emission.
If the source of light is brought near the photocells and placed 50 cm away from them, then the intensity of radiation will increase. This does not affect the energy of the radiation. Hence, the result will be the same as before. However, the photoelectrons emitted from Na and K will increase in proportion to intensity.
संबंधित प्रश्न
Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is −1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity (∼105 W m−2) red light of wavelength 6328 Å produced by a He-Ne laser?
Can a photon be deflected by an electric field? Or by a magnetic field?
The threshold wavelength of a metal is λ0. Light of wavelength slightly less than λ0 is incident on an insulated plate made of this metal. It is found that photoelectrons are emitted for some time and after that the emission stops. Explain.
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A photon of energy hv is absorbed by a free electron of a metal with work-function hv − φ.
The collector plate in an experiment on photoelectric effect is kept vertically above the emitter plate. A light source is put on and a saturation photocurrent is recorded. An electric field is switched on that has a vertically downward direction.
A 100 W light bulb is placed at the centre of a spherical chamber of radius 20 cm. Assume that 60% of the energy supplied to the bulb is converted into light and that the surface of the chamber is perfectly absorbing. Find the pressure exerted by the light on the surface of the chamber.
(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 work function of a metal is 2.5 × 10−19 J. (a) Find the threshold frequency for photoelectric emission. (b) If the metal is exposed to a light beam of frequency 6.0 × 1014 Hz, what will be the stopping potential?
(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 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)
A small piece of cesium metal (φ = 1.9 eV) is kept at a distance of 20 cm from a large metal plate with a charge density of 1.0 × 10−9 C m−2 on the surface facing the cesium piece. A monochromatic light of wavelength 400 nm is incident on the cesium piece. Find the minimum and maximum kinetic energy of the photoelectrons reaching the large metal plate. Neglect any change in electric field due to the small piece of cesium present.
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
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?
In the case of photoelectric effect experiment, explain the following facts, giving reasons.
The photoelectric current increases with increase of intensity of incident light.
Define the term: stopping potential in the photoelectric effect.
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.
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.
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.
