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What is so special about the combination e/m? Why do we not simply talk of e and m separately?

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प्रश्न

What is so special about the combination e/m? Why do we not simply talk of e and m separately?

संक्षेप में उत्तर
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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.

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अध्याय 11: Dual Nature of Radiation and Matter - Exercise [पृष्ठ ४११]

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एनसीईआरटी Physics Part I and II [English] Class 12
अध्याय 11 Dual Nature of Radiation and Matter
Exercise | Q 11.37 (b) | पृष्ठ ४११

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.


(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?


If light of wavelength 412.5 nm is incident on each of the metals given below, which ones will show photoelectric emission and why?

Metal Work Function (eV)
Na 1.92
K 2.15
Ca 3.20
Mo 4.17

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 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)


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. 


The stopping potential in an experiment on photoelectric effect is 1.5V. What is the maximum kinetic energy of the photoelectrons emitted? Calculate in Joules.


In the experimental set up for studying photoelectric effect, if keeping the frequency of the incident radiation and the accelerating potential fixed, the intensity of light is varied, then ______.


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 ______.


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 ______.


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In photoelectric effect, the photoelectric current


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.


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