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

Correct answer is A

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

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