An electron of charge 1.6 x 10-19C and mass 9.1 x 10-31kg is accelerated between two metal plates with a velocity of 4 x 107ms-1, the potential difference between the plates is
-
4.55 x 103V
- 4.55 x 102V
- 9.10 x 101V
- 4.55 x 101V
The correct answer is: A
Explanation
When a charged particle is accelerated across a region of p.d. V Volt, the work done is the product of the charge Q and the p.d. Vi.e. Work done = Q x V
But this work done is equal to the K.E. imparted to the charge given by K.E = 1/2MV2, where V is the velocity of the charge = Q x V = 1/2MV2
∴ p.d V = (1/2MV2)/Q
= 1/2 x 9.1-31 x [4 x 107]2/1.6 x 10-19
= 8 x 9.1 x 102/1.6
= 4550V
= 4.55 x 103V
There is an explanation video available .