Advertisements
Advertisements
प्रश्न
A silver ball of radius 4.8 cm is suspended by a thread in a vacuum chamber. Ultraviolet light of wavelength 200 nm is incident on the ball for some time during which light energy of 1.0 × 10−7 J falls on the surface. Assuming that on average, one photon out of every ten thousand is able to eject a photoelectron, find the electric potential at the surface of the ball, assuming zero potential at infinity. What is the potential at the centre of the ball?
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
Advertisements
उत्तर
Given:-
Radius of the silver ball, r = 4.8 cm
Wavelength of the ultra violet light, λ = 200 nm = 2 × 10−7 m
Total energy of light, E = 1.0 × 10−7 J
We are given that one photon out of every ten thousand is able to eject a photoelectron.
Energy of one photon,
`E^' = (hc)/lambda`,
where h = Planck's constant
c = speed of light
`lambda` = wavelength of light
On substituting the respective values in the above formula, we get :
`E^' = (6.63 xx 10^-34 xx 3 xx 10^8)/(2 xx 10^-7)`
`=9.945 xx 10^-19`
Number of photons,
`n = E/E^' = (1 xx 10^-7)/(9.945 xx 10^-19) = 1 xx 10^11`
Number of photoelectrons
= `(1 xx 10^11)/10^4 = 1 xx 10^7`
The amount of positive charge developed due to the outgoing electrons,
`q = 1 xx 10^7 xx 1.6 xx 10^-19`
`= 1.6 xx 10^-12 C`
Potential developed at the centre as well as on surface,
`V = (Kq)/r`,
where K = `1/(4piε_0)`
`therefore V = (9 xx 10^9 xx 1.6 xx 10^-12)/(4.8 xx 10^-2) = 0.3 V`
Potential inside the silver ball remains constant. Therefore, potential at the centre of the sphere is 0.3 V.
APPEARS IN
संबंधित प्रश्न
The threshold frequency for a certain metal is 3.3 × 1014 Hz. If light of frequency 8.2 × 1014 Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.
Light of frequency 7.21 × 1014 Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 105 m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
An electron gun with its collector at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (∼10−2 mm of Hg). A magnetic field of 2.83 × 10−4 T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the ‘fine beam tube’ method. Determine e/m from the data.
What is so special about the combination e/m? Why do we not simply talk of e and m separately?
Define the term "cut off frequency" in photoelectric emission. The threshod frequency of a metal is f. When the light of frequency 2f is incident on the metal plate, the maximum velocity of photo-electrons is v1. When the frequency of the incident radiation is increased to 5f, the maximum velocity of phto-electrons is v2. Find the ratio v1 : v2.
Visible light has wavelengths in the range of 400 nm to 780 nm. Calculate the range of energy of the photons of visible light.
(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 neutral particles are kept 1 m apart. Suppose by some mechanism some charge is transferred from one particle to the other and the electric potential energy lost is completely converted into a photon. Calculate the longest and the next smaller wavelength of the photon possible.
(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, light of wavelength 400 nm is incident on a cesium plate at the rate of 5.0 W. The potential of the collector plate is made sufficiently positive with respect to the emitter, so that the current reaches its saturation value. Assuming that on average, one out of every 106 photons is able to eject a photoelectron, find the photocurrent in the circuit.
A horizontal cesium plate (φ = 1.9 eV) is moved vertically downward at a constant speed v in a room full of radiation of wavelength 250 nm and above. What should be the minimum value of v so that the vertically-upward component of velocity is non-positive for each photoelectron?
(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 the arrangement shown in the figure, y = 1.0 mm, d = 0.24 mm and D = 1.2 m. The work function of the material of the emitter is 2.2 eV. Find the stopping potential V needed to stop the photocurrent.

(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
Work function of aluminium is 4.2 eV. If two photons each of energy 2.5 eV are incident on its surface, will the emission of electrons take place? Justify your answer.
In the case of a photo electric effect experiment, explain the following facts, giving reasons.
The wave theory of light could not explain the existence of the threshold frequency.
For a given frequency of light and a positive plate potential in the set up below, If the intensity of light is increased then ______.

In various experiments on photo electricity, the stopping potential for a given frequency of the incident radiation is ______.
The electromagnetic theory of light failed to explain ______.
An increase in the intensity of the radiation causing photo-electric emission from a surface does not affect the maximum K.E. of the photoelectrons. Explain.
