Advertisements
Advertisements
प्रश्न
What is so special about the combination e/m? Why do we not simply talk of e and m separately?
Advertisements
उत्तर
The basic relations for electric field and magnetic field are `("eV" = 1/2 "mv"^2)` and `("eBv" = ("mv"^2)/"r")` respectively
These relations include e (electric charge), v (velocity), m (mass), V (potential), r(radius), and B (magnetic field). These relations give the value of velocity of an electron as `["v" = sqrt(2"V"("e"/"m"))]` and `["v" = "Br"("e"/"m")]` respectively.
It can be observed from these relations that the dynamics of an electron are determined not by e and m separately, but by the ratio e/m.
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?
(a) A monoenergetic electron beam with electron speed of 5.20 × 106 m s−1 is subject to a magnetic field of 1.30 × 10−4 T normal to the beam velocity. What is the a radius of the circle traced by the beam, given e/m for electron equals 1.76 × 1011 C kg−1?
(b) Is the formula you employ in (a) valid for calculating the radius of the path of a 20 MeV electron beam? If not, in what way is it modified?
Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified?
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)
A photographic film is coated with a silver bromide layer. When light falls on this film, silver bromide molecules dissociate and the film records the light there. A minimum of 0.6 eV is needed to dissociate a silver bromide molecule. Find the maximum wavelength of light that can be recorded by the film.
(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)
Plot a graph to show the variation of stopping potential with frequency of incident radiation in relation to photoelectric effect.
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.
In various experiments on photo electricity, the stopping potential for a given frequency of the incident radiation is ______.
When a beam of 10.6 eV photons of intensity 2.0 W/m2 falls on a platinum surface of area 1.0 × 10-4 m2, only 53% of the incident photons eject photoelectrons. The number of photoelectrons emitted per second is ______.
The electromagnetic theory of light failed to explain ______.
Cathode rays can be deflected by
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.
