#### 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, *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 (*mv*^{2}/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))`

#### APPEARS IN

#### Related Questions VIEW ALL [5]

If light of wavelength 412.5 nm is incident on each of the metals given below, which ones will show photoelectric emission and why?

Metal |
Work Function (eV) |

Na | 1.92 |

K | 2.15 |

Ca | 3.20 |

Mo | 4.17 |