मराठी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

(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. - Physics

Advertisements
Advertisements

प्रश्न

(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?

संख्यात्मक
Advertisements

उत्तर

(a) Speed of an electron, v = 5.20 × 106 m/s

Magnetic field experienced by the electron, B = 1.30 × 10−4 T

Specific charge of an electron, e/m = 1.76 × 1011 C kg−1

Where,

e = Charge on the electron = 1.6 × 10−19 C

m = Mass of the electron = 9.1 × 10−31 kg−1

The force exerted on the electron is given as:

`"F" = "e"|vec"v" xx vec"B"|`

= evB sin θ

θ = Angle between the magnetic field and the beam velocity

The magnetic field is normal to the direction of beam.

∴ θ = 90°

F = evB .............(1)

The beam traces a circular path of radius, r. It is the magnetic field, due to its bending nature, that provides the centripetal force `("F" = ("mv"^2)/"r")` for the beam.

Hence, equation (1) reduces to:

`"evB" = ("mv"^2)/"r"`

∴ r = `"mv"/("eB") = "v"/(("e"/"m")"B")`

= `(5.20 xx 10^6) /((1.76 xx 10^11) xx 1.30 xx    10^(-4))`

= 0.227 m

= 22.7 cm

Therefore, the radius of the circular path is 22.7 cm.

(b) Energy of the electron beam, E = 20 MeV = 20 × 106 × 1.6 × 10−19 J

The energy of the electron is given as:

`"E" = 1/2 "mv"^2`

∴ v = `((2"E")/"m")^(1/2)`

= `sqrt((2xx20xx10^6xx 1.6 xx 10^(-19))/(9.1 xx 10^(-31)))`

= 2.652 × 109 m/s

This result is incorrect 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.

When very high speeds are concerned, the relativistic domain comes into consideration.

In the relativistic domain, mass is given as:

`"m" = "m"_0 [1 - "v"^2/"c"^2]^(1/2)`

Where,

`"m"_0` = Mass of the particle at rest

Hence, the radius of the circular path is given as:

r = `"mv"/"eB"`

= `("m"_0"v")/("eB"sqrt(("c"^2 - "v"^2)/"c"^2))`

shaalaa.com
Photoelectric Effect and Wave Theory of Light
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?

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?


 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?


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

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)


A light beam of wavelength 400 nm is incident on a metal plate of work function 2.2 eV. (a) A particular electron absorbs a photon and makes two collisions before coming out of the metal. Assuming that 10% of the extra energy is lost to the metal in each collision, find the kinetic energy of this electron as it comes out of the metal. (b) Under the same assumptions, find the maximum number of collisions the electron can suffer before it becomes unable to come out of the metal.


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. 


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 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 ______.


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 ______.


Cathode rays can be deflected by


In photoelectric effect, the photoelectric current


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×