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
A
4.55 x 103V
B
4.55 x 102V
C
9.10 x 101V
D
4.55 x 101V
Verified Correct Answer: A.
Physical Principles & Numerical Solution
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. V
i.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
Source: Myschool Past Questions Archive (Record #713)