Q. 213.9( 8 Votes )

# (a) A monoenerget

Answer :

Given:

Speed of electron, v = 5.20 × 106 ms-1

Magnetic field strength normal to the beam = 1.30 × 10–4 T

Charge to mass ratio (e/m) = 1.76 × 1011 C Kg-1

(a) Force applied by magnetic field,

F = e|v × B|

F = e × v × B × sinθ …(1)

Where,

e = Charge on electron

v = velocity of particle

B = magnetic field strength

θ = angle between Magnetic field and velocity

Since the electron traces a circular path, we can use the following equation of centrifugal force:

F = …(2)

Where,

m = mass of particle

v = velocity of particle

r = radius of circle traced

By equating (1) and (2), we get,

evB × sin90° = r = r = …(3)

r = r = 22.7 6 cm

(b) Energy of the electron beam, E = 20 MeV

E = 20 × 106 × 1.6 × 10-19J

The energy of an electron is given by:

E = From the above equation, we can write,

v = …(4)

putting the value in equation (4)

v = v = 2.6 × 109 ms-1

This result is incorrect because the speed of any massive object can’t exceed the speed of light (i.e. 3 × 108 ms-1). We can’t use equation (4) in the case where the speed is relativistic:

At relativistic speeds, mass is given by,

m = Where,

m = relativistic mass

m0 = rest mass

v = velocity of particle

c = speed of light

the radius of the path traced is given by,

r = …(3)

By substituting the value of relativistic mass in equation (3) we get,

r = …(4)

By using equation (4) we can find the radius traced by electrons moving at relativistic speed.

By substituting value of relativistic mass in equation

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