Q. 213.9( 8 Votes )

# (a) A monoenerget

Answer :

Given:

Speed of electron, v = 5.20 × 10^{6} ms^{-1}

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

Charge to mass ratio (**e/m**) = 1.76 × 10^{11} 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 × 10^{6} × 1.6 × 10^{-19}J

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 × 10^{9} ms^{-1}

This result is incorrect because the speed of any massive object can’t exceed the speed of light (i.e. 3 × 10^{8} 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

m_{0} = 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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