An electron moving with the speed `5xx10^(6)` per sec is shot parallel to the electric field of intensity `1xx10^(3)N//C`. Field is responsible for the retardation of motion of electron. Now evaluate the distance travelled by the electron before coming to rest for an instant (mass of `e= 9xx10^(-31)Kg` charge `= 1.6xx10^(-19)C)`

Options

A. 7 m
B. 0.7 mm
C. 7 cm
D. 0.7 cm

Answer

C

Explanation

`S=u^2/(2a) = (5 xx10^(6))^(2)/(2xx(qE)/m)=(25 xx10^(12) xx9 xx10^(-31))/(2xx1.6 xx10^(-19) xx10^3)`
`= 0.07m = 7 cm.`