Q - 18

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 , v_y = 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  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