Advertisements
Advertisements
प्रश्न
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.
विकल्प
The photocurrent will increase.
The kinetic energy of the electrons will increase.
The stopping potential will decrease.
The threshold wavelength will increase.
Advertisements
उत्तर
The kinetic energy of the electrons will increase.
As there is no effect of electric field on the number of photons emitted, the photoelectric current will remain same. Hence, option (a) is incorrect.
When an electric field is applied, then electric force will act on the electron moving opposite the direction of electric field, which will increase the kinetic energy of the electron. Hence, option (b) is correct.
As the kinetic energy of the electron is increasing, its stopping potential will increase. Hence, option (c) is incorrect.
Threshold wavelength is the characteristic property of the metal and will not change. Hence, (d) is incorrect.
APPEARS IN
संबंधित प्रश्न
Monochromatic radiation of wavelength 640.2 nm (1 nm = 10−9 m) from a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.
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?
If an electron has a wavelength, does it also have a colour?
Light of wavelength λ falls on a metal with work-function hc/λ0. Photoelectric effect will take place only if
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)
Find the maximum kinetic energy of the photoelectrons ejected when light of wavelength 350 nm is incident on a cesium surface. Work function of cesium = 1.9 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)
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 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)
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.
(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)
The figure is the plot of stopping potential versus the frequency of the light used in an experiment on photoelectric effect. Find (a) the ratio h/e and (b) the work function.
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 thin target (10–2 cm square, 10–3 m thickness) of sodium, which produces a photocurrent of 100 µA when a light of intensity 100W/m2 (λ = 660 nm) falls on it. Find the probability that a photoelectron is produced when a photons strikes a sodium atom. [Take density of Na = 0.97 kg/m3].
The work function for a metal surface is 4.14 eV. The threshold wavelength for this metal surface is ______.
The graph shows the variation of photocurrent for a photosensitive metal
- What does X and A on the horizontal axis represent?
- Draw this graph for three different values of frequencies of incident radiation ʋ1, ʋ2 and ʋ3 (ʋ3 > ʋ2 > ʋ1) for the same intensity.
- Draw this graph for three different values of intensities of incident radiation I1, I2 and I3 (I3 > I2 > I1) having the same frequency.
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.
