# Physics Class 12 NCERT Solutions: Chapter 11 Dual Nature of Radiation and Matter Part 9 (For CBSE, ICSE, IAS, NET, NRA 2022)

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Q: 21. (A) A mono-energetic electron beam with electron speed of is subject to a magnetic field of normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals .

(B) Is the formula you employ in (a) valid for calculating radius of the path of a electron beam? If not, in what way is it modified?

[Note: Exercises 20 (b) and 21 (b) (Reference: Chapter 11 Dual Nature of Radiation and Matter Part 8) take you to relativistic mechanics which is beyond the scope of this book. They have been inserted here simply to emphasise the point that the formulas you use in part (a) of the exercises are not valid at very high speeds or energies. See answers at the end to know what ‘very high speed or energy’ means.]

Answer:

(A) Speed of an electron,

Magnetic field experienced by the electron,

Specific charge of an electron,

Where,

Charge on the electron

Mass of the electron

The force exerted on the electron is given as:

Angle between the magnetic field and the beam velocity

The magnetic field is normal to the direction of beam.

The beam traces a circular path of radius, . It is the magnetic field, due to its bending nature, that provides the centripetal force for the beam.

Hence, equation (1) reduces to:

Therefore, the radius of the circular path is .

(B) Energy of the electron beam,

The energy of the electron is given as:

This result is incorrect because nothing can move faster than light. In the above formula, the expression for energy can only be used in the non-relativistic limit, i.e.. , for

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

In the relativistic domain, mass is given as:

Where,

Mass of the particle at rest

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