Question

3. (a) Calculate the speed of a proton that is accelerated from rest through an electric potential difference of 136 V.
m/s

(b) Calculate the speed of an electron that is accelerated through the same potential difference.
m/s

Answer 1

we know that kinetic energy for a particle is gievn by the equation

KE = mv²/2

where, m =mass of particle

v= =velocity of the particle

given that an electric potential difference of 136 V, since both proton snd elelctron have same electric charge, with an opposite sign, so both the particles will gain 136 V of kinetic energy

converting eV to J using the relation

1 eV = 1.6x 10^-19 J

hence , 136 eV = 136x 1.6 x 10^-19 J = 217.6 x 10^-19 J

a) for proton , KE = 217.6 x 10^-19 J, m = 1.67 x 10^-27 kg

therefore KE = mv² / 2

or v² = 2(KE)/m = 2(217.6 x 10^-19) / ( 1.67 x 10^-27) = 260.6 x 10⁸ m/s

or v = 16.14 x 10⁴ m/s = 1.614 x 10⁵ m/s

b) similarly for electron, KE = 217.6 x 10^-19, m = 9.1 x 10^-31 kg

therefore

v² = 2(KE)/m = 2(217.6 x 10^-19) / ( 9.1 x 10^-31) = 47.8 x 10¹² m/s

or v = 6.91 x 10⁶ m/s