Advertisement Remove all ads

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? - Physics

 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?

Advertisement Remove all ads


Energy of the electron beam, = 20 MeV = `20 xx 10^6 xx 1.6 xx 10^(-19) J`

The energy of the electron is given as:

`E = 1/2 mv^2`

`:. v = (2E/m)^(1/2)`

`= sqrt((2xx20xx10^6xx 1.6 xx 10^(-19))/(9.1 xx 10^(-31)))  = 2.652 xx 10^(9) "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)`


`m_0` = Mass of the particle at rest

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

`r = mv/eB`

`= (m_0v)/(eBsqrt((c^2-v^2)/c^2))`


[Note: Exercises 11.20(b) and 11.21(b) 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.]

Concept: Photoelectric Effect and Wave Theory of Light
  Is there an error in this question or solution?
Advertisement Remove all ads


NCERT Class 12 Physics Textbook
Chapter 11 Dual Nature of Radiation and Matter
Q 21.2 | Page 409
Advertisement Remove all ads
Advertisement Remove all ads

View all notifications

      Forgot password?
View in app×