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 ? , [ Note :
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 ? ,
[ Note : Exercises 20 (b) and (21) b take you to relativistic mechanics which is beyond the scope of this book . They have been inserted here simply to emphasise the point that the formulas you use in part (a) of the exercises are not valid at very high speeds or energies . See answers at the end to know what very high speed or energy means .]
1 Answers
Given , E = 20 MeV = ` 20 xx 1.6 xx 10^(-13)` J, `m_(e) = 9.1 xx10^(-31)` kg
` E = 1/2 mv^(2)`
` rArr v = sqrt((2E)/m) = sqrt((2 xx20 xx1.6 xx10^(-13))/(9.1xx10^(-32))) :. v = 2.67 xx10^(9) ` m/s
As `upsilon gt C,` the formula used in (a) `r=(mv)/(eB)` is not valid for calculating the radius of path of `2theta` MeV electron beam because electron with such a high energy has velocity in relatistic domain i.e., comparable with the velocity of light and the mass varies with the increase in velocity but we have taken it as constant.
`thereforem=(m_(0))/(sqrt(1-(v^(2))/(c^(2))))rArr " Thus, the modified formula will be r"=(muv)/(eB)[(m_(0))/(sqrt(1-(v^(2))/(c^(2))))](v)/(eB)`