CBSE (Science) Class 12CBSE
Share
Notifications

View all notifications
Books Shortlist
Your shortlist is empty

Solution for 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? - CBSE (Science) Class 12 - Physics

Login
Create free account


      Forgot password?

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?

Solution

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)`

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))`

  Is there an error in this question or solution?

APPEARS IN

 NCERT Physics Textbook for Class 12 Part 2 (with solutions)
Chapter 11: Dual Nature of Radiation and Matter
Q: 21.2 | Page no. 409
Solution for 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? concept: Photoelectric Effect and Wave Theory of Light. For the course CBSE (Science)
S
View in app×