Question

 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?

Answer 1

Energy of the electron beam, E = 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)`

Where,

`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))`