Products
Rewards 
from HOLOOLY

We are determined to provide the latest solutions related to all subjects FREE of charge!

Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

HOLOOLY 
BUSINESS MANAGER

Advertise your business, and reach millions of students around the world.

HOLOOLY 
TABLES

All the data tables that you may search for.

HOLOOLY 
ARABIA

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

HOLOOLY 
TEXTBOOKS

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

HOLOOLY 
HELP DESK

Need Help? We got you covered.

Chapter 3

Q. 3.4

In a parallel diode, the cathode and anode are spaced 5 mm apart and the anode is kept at 200 V d.c. with respect to cathode. Calculate the velocity and the distance travelled by an electron after a time of 0.5 nsec, when (a) the initial velocity of an electron is zero and (b) the initial velocity is 2 × 10^{6} m/sec in the direction towards the anode.

Step-by-Step

Verified Solution

Given V = 200 V, d = 5 mm, Time of travel by an electron, t = 0.5 nsec.

Therefore,          E=\frac{V}{d}=\frac{200}{5 \times 10^{-3} } =40\times \frac{10^{3}V }{m}

a_{x}=\frac{qE}{m}=\frac{1.602\times 10^{-19}\times 40\times 10^{3} }{9.1\times 10^{-31} } =70.33\times 10^{14}m/\sec ^{2}

We know that

Velocity of an electron,       v_{x} = v_{ox} + a_{x}t

Distance travelled by an electron, x = v_{ox} + 0.5at^{2}

(a) When the initial velocity of an electron is zero

v_{x} = v_{ox} + a_{x}t = 0 + 70.33 × 10^{14} × 0.5 × 10^{–9} = 3.5165 × 10^{6} m/sec

x = v_{ox}t + 0.5 at^{2} = 0 + 0.5 × 70.33 × 10^{14} × (0.5 × 10^{–9})^{2} = 0.879 × 10^{–3} m

(b) When the initial velocity of an electron is v_{ox} = 2\times \frac{10^{6}m }{sec}

v_{x} = v_{ox} + a_{x}t = 2 × 10^{6} + 70.33 × 10^{14} × 0.5 × 10^{–9} = 5.51 × 10^{6} m/sec

x = v_{ox}t + 0.5 at^{2} = 2 × 10^{6} × 0.5 × 10^{–9} + 0.5 × 70.33 × 10^{14} × (0.5 × 10^{–9})^{2}

= 1.879 × 10^{–3} m