Advertisements
Advertisements
प्रश्न
(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?
उत्तर
(a) Potential difference across the evacuated tube, V = 500 V
Specific charge of an electron, e/m = 1.76 × 1011 C kg−1
The speed of each emitted electron is given by the relation for kinetic energy as:
`"KE" = 1/2 "mv"^2 = "eV"`
∴ v = `((2"eV")/"m")^(1/2) = (2"V" xx "e"/"m")^(1/2)`
= `(2 xx 500 xx 1.76 xx 10^11)^(1/2)`
= 1.327 × 107 m/s
Therefore, the speed of each emitted electron is 1.327 × 107 m/s.
(b) Potential of the anode, V = 10 MV = 10 × 106 V
The speed of each electron is given as:
v = `(2"V" "e"/"m")^(1/2)`
= `(2 xx 10^7 xx 1.76 xx 10^11)^(1/2)`
= 1.88 × 109 m/s
This result is wrong because nothing can move faster than light. In the above formula, the expression (mv2/2) for energy can only be used in the non-relativistic limit, i.e., for v << c.
For very high-speed problems, relativistic equations must be considered for solving them. In the relativistic limit, the total energy is given as:
E = mc2
Where,
m = Relativistic mass
= `"m"_0 (1 - "v"^2/"c"^2)^(1/2)`
m0 = Mass of the particle at rest
Kinetic energy is given as:
K = mc2 − m0c2
APPEARS IN
संबंधित प्रश्न
Define the term 'intensity of radiation' in terms of photon picture of light.
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?
Is it always true that for two sources of equal intensity, the number of photons emitted in a given time are equal?
Can a photon be deflected by an electric field? Or by a magnetic field?
Should the energy of a photon be called its kinetic energy or its internal energy?
Planck's constant has the same dimensions as
The work function of a metal is hv0. Light of frequency v falls on this metal. Photoelectric effect will take place only if
If the frequency of light in a photoelectric experiment is doubled, the stopping potential will ______.
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
If the wavelength of light in an experiment on photoelectric effect is doubled,
(a) photoelectric emission will not take place
(b) photoelectric emission may or may not take place
(c) the stopping potential will increase
(d) the stopping potential will decrease
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 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)
When a metal plate is exposed to a monochromatic beam of light of wavelength 400 nm, a negative potential of 1.1 V is needed to stop the photo current. Find the threshold wavelength for the metal.
(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.
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)
In the case of photoelectric effect experiment, explain the following facts, giving reasons.
The photoelectric current increases with increase of intensity of incident light.
In photoelectric effect the photo current ______.
If photons of ultraviolet light of energy 12 eV are incident on a metal surface of work function of 4 eV, then the stopping potential (in eV) will be :
