Advertisements
Advertisements
Question
(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?
Advertisements
Solution
(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))`
RELATED QUESTIONS
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?
What is so special about the combination e/m? Why do we not simply talk of e and m separately?
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 |
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)
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 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.
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)
In an experiment on photoelectric effect, the emitter and the collector plates are placed at a separation of 10 cm and are connected through an ammeter without any cell. A magnetic field B exists parallel to the plates. The work function of the emitter is 2.39 eV and the light incident on it has wavelengths between 400 nm and 600 nm. Find the minimum value of B for which the current registered by the ammeter is zero. Neglect any effect of space charge.
Plot a graph to show the variation of stopping potential with frequency of incident radiation in relation to photoelectric effect.
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.
Answer the following question.
Why is the wave theory of electromagnetic radiation not able to explain the photoelectric effect? How does a photon picture resolve this problem?
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 ______.
