Q - 18

Asked by majethiyarishat9566.12sdatl | 10th Nov, 2020, 09:42: PM

Expert Answer:

Energy of electron before entering electric field = (1/2) m v2 = 1000 J
 
where m is mass of electron and v is speed of electron before entering electric field 
 
speed v of electron before entering electric field = { 2000 / m }1/2 = { 2000 / ( 9.1 × 10-31 ) }1/2 = 4.69 × 1016 m/s
 
Time to cross plate of length 2 cm = distance /speed = 0.02 / ( 4.69 × 1016 ) = 4.264 × 10-19 s
 
Force F acting on electron , F = e × E 
 
where e = 1.602 × 10-19 C , is charge on electron and E = 3  V/cm = 300 V/m , intensity of electric field .
 
acceleration = Force / mass = ( 1.602 × 10-19 × 300 ) / ( 9.1 × 10-31 )  = 5.281 × 1013 m/s2 
 
vertical component of velocity when electron leaves the plate , vy = a × t = 5.281 × 1013 × 4.264 × 10-19 = 2.25 × 10-5 m/s
 
vertical distance travelled by  electron when it leaves the plates,  h = (1/2) a t2 = 0.5 × 5.281 × 1013 × ( 4.264 × 10-19 )2 
 
vertical distance travelled by  electron when it leaves the plates,  h = 4.8 × 10-24 m
 
Question (a) :-
 
horizontal componenet of velocity is greater than vertical componenet of velocity  by order 1021  
 
Hence velocity vector of electron when it leaves the plate  has magntiude and direction of horizontal component velocity
 
velocity vector of electron when it leaves the plate = 4.69 × 1016 begin mathsize 14px style i with hat on top end style m/s
 
Question (b)
 
angle θ of deflection is given as , tanθ = ( vertical distance travelled by  electron ) / plate length
 
tanθ = ( 4.8 × 10-24 ) / 0.02 = 2.4 × 10-22 
 
tanθ is so small, we get angle of deflection = 2.4 × 10-22 rad
 
Question (c) 
 
velocity is least affected and deflection is negligibly small , Hence electron almost travel along x-axis

Answered by Thiyagarajan K | 10th Nov, 2020, 11:49: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.