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Solution - 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 (Commerce) Class 12 - Physics

ConceptPhotoelectric Effect and Wave Theory of Light

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

Reference Material

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: null - Photoelectric Effect and Wave Theory of Light. For the courses CBSE (Commerce), CBSE (Arts), CBSE (Science)
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