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NCERT Class XII Chapter
Dual Nature of Radiation and Matter
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Question : 21
Total: 37
(a) A monoenergetic electron beam with electron speed of 5.20 × 106 m s–1 is subjected to a magnetic field of 1.30 × 10–4 T, normal to the beam velocity. What is the 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 radius of the path of a20 MeV electron beam? If not, in what way is it modified?
Solution:  
(a) Here v = 5.20 × 106ms–1,
B = 1.30 × 10–4 T
Specific charge,
e
m
 = 1.76 × 1011 C kg−1 , θ = 90°
The normal magnetic field provides necessary centripetal force to the electron beam so that it can follow a circular path. Thus Force on an electron = Centripetal force due to magnetic field on an electron
or evB sin 90° =
mv2
r
 or r =
mv
eB
 =
v
(e∕m)B
=
5.20×10
1.7×1011×1.30×10−4
 m = 0.227 m = 22.7 cm
(b) r =
mv
eB
The formula for radius of circular path is not valid at very high energies because such high energy electrons have velocities comparable to the speed of light.
In such situation we use relativistic formula for mass of electron.
m = m0√1−
v2
c2
 , where m0 is rest mass
Radius r =
mv
eB
 =
m0
√1−
v2
c2
 (
v
eB
)
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