Question

In a cathode ray tube, electrons are accelerated from rest by a constant electric force of magnitude 6.40 × 10⁻¹⁷ N during the first 2.60 cm of the tube’s length; then they move at essentially constant velocity another 45.0 cm before hitting the screen.

(a)Find the speed of the electrons when they hit the screen. ___m/s

(b)How long does it take them to travel the length of the tube? ___ns

Answer 1

Step 1: Using Newton's 2nd law:

F_net = m*a

a = acceleration of electrons = F_net/m

F_net = net electric force on electron = 6.40*10^-17 N

m = mass of electron = 9.1*10^-31 kg

So,

a = 6.40*10^-17/(9.1*10^-31) = 7.033*10^13 m/s^2

Step 2: Using 3rd kinematic equation find speed of electron after it travels 2.60 cm

V^2 = U^2 + 2*a*d1

V = final speed after it travels 2.60 cm = ?

U = Initial speed = 0 m/s

So,

V = sqrt (0^2 + 2*7.033*10^13*2.60*10^-2) = 1912370.26 m/s

V = 1.912*10^6 m/s

Now after that electron travels at constant speed, So

Speed of electrons when they hit the screen = 1.912*10^6 m/s

Part B.

Time taken by electron to travel 2.60 cm will be, Using 1st kinematic equation

V = U + a*t1

t1 = (V - U)/a

t1 = (1912370.26 - 0)/(7.033*10^13) = 2.72*10^-8 sec

Now after that time taken by electron to travel 45.0 cm at constant speed will be:

t2 = distance/Speed

t2 = 0.45/1912370.26 = 23.53*10^-8 sec

So total time will be:

t = t1 + t2

t = 2.72*10^-8 + 23.53*10^-8

t = 262.5*10^-9 sec

t = 262.5 ns

Let me know if you've any query.