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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? - Photoelectric Effect and Wave Theory of Light

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